AS Differentiation
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stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.
"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.
So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$
This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.
derivatives
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
$endgroup$
add a comment |
$begingroup$
stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.
"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.
So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$
This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.
derivatives
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
$endgroup$
1
$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36
add a comment |
$begingroup$
stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.
"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.
So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$
This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.
derivatives
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
$endgroup$
stuck on this question, I've tried it many times but to no avail. There's a mark scheme that has no detailed working.
"A closed cylinder is such that its surface area is 50π cm$^2$. Calculate a) the radius of the cylinder that gives the max volume. b) the maximum volume and prove it is a maximum.
So starting off, I used the formula $A$ = 2πrh + 2πr$^2$, in which 50π = 2πrh + 2πr$^2$. What I did next was to rearrange for r, in which I got 50π = 2πr(h+r), which simplifies more to 25 = r(h+r), which goes to $$h = frac{25}{r}-r$$
This can then be subbed back into the formula to find r, but this is where I am completely muddled. Please save me from this 67 question hell, thank you.
derivatives
derivatives
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
asked Jan 7 at 20:32
Jia Xuan NgJia Xuan Ng
101
101
1
$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36
add a comment |
1
$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36
1
1
$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36
$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36
add a comment |
2 Answers
2
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oldest
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$$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$
From here, you correctly solved for $h$ in terms of $r$:
$$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$
You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.
$$V = bh = pi r^2h$$
Plugging in $(1)$, you get
$$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$
Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:
$$frac{dV}{dr} = 25pi-3pi r^2$$
$$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$
Can you continue from here?
Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.
answered Jan 7 at 21:00
KM101KM101
6,0251525
$endgroup$
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
|
show 1 more comment
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Hint:
The volume of a Cylinder $V=r^2pi h$
Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$
Now look for the maximum of that Function
answered Jan 7 at 20:42
A. PA. P
1085
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$
From here, you correctly solved for $h$ in terms of $r$:
$$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$
You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.
$$V = bh = pi r^2h$$
Plugging in $(1)$, you get
$$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$
Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:
$$frac{dV}{dr} = 25pi-3pi r^2$$
$$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$
Can you continue from here?
Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.
answered Jan 7 at 21:00
KM101KM101
6,0251525
$endgroup$
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
|
show 1 more comment
$begingroup$
$$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$
From here, you correctly solved for $h$ in terms of $r$:
$$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$
You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.
$$V = bh = pi r^2h$$
Plugging in $(1)$, you get
$$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$
Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:
$$frac{dV}{dr} = 25pi-3pi r^2$$
$$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$
Can you continue from here?
Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.
answered Jan 7 at 21:00
KM101KM101
6,0251525
$endgroup$
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
|
show 1 more comment
$begingroup$
$$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$
From here, you correctly solved for $h$ in terms of $r$:
$$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$
You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.
$$V = bh = pi r^2h$$
Plugging in $(1)$, you get
$$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$
Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:
$$frac{dV}{dr} = 25pi-3pi r^2$$
$$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$
Can you continue from here?
Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.
answered Jan 7 at 21:00
KM101KM101
6,0251525
$endgroup$
$$A_{s} = 2pi rh + 2pi r^2 implies 50pi = 2pi rh+2pi r^2$$
From here, you correctly solved for $h$ in terms of $r$:
$$25pi = pi rh+pi r^2 implies 25 = rh+r^2 implies 25 = r(h+r) implies color{blue}{h = frac{25}{r}-r} tag{1}$$
You have to find the radius $r$ which gives the maximum volume, so you must use the formula for the volume of a cylinder.
$$V = bh = pi r^2h$$
Plugging in $(1)$, you get
$$V = pi r^2left(frac{25}{r}-rright) = 25pi r-pi r^3$$
Treat $V$ as $f(r)$, a function of $r$, take the derivative, and set it to $0$:
$$frac{dV}{dr} = 25pi-3pi r^2$$
$$frac{dV}{dr} = 0 implies 0 = 25pi-3pi r^2$$
Can you continue from here?
Edit: You can also use the fact that $frac{dV}{dr}$ is a quadratic with a negative coefficient for the leading term, so the point $r = -frac{b}{2a}$ must be a maximum.
answered Jan 7 at 21:00
KM101KM101
6,0251525
edited Jan 7 at 22:57
answered Jan 7 at 21:00
KM101KM101
6,0251525
answered Jan 7 at 21:00
KM101KM101
6,0251525
answered Jan 7 at 21:00
KM101KM101
6,0251525
6,0251525
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
|
show 1 more comment
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
I have tried it multiple times, by doing second derivative. I always end up with 6pi r, which confuses me. I'll try it again tommorow and see if it works, thank you very much.
$endgroup$
– Jia Xuan Ng
Jan 7 at 22:41
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
You don’t even need derivatives. The quadratic has a negative coefficient for the quadratic term, that proves that the point is a maximum. Also, you have $frac{d^2V}{dr^2} = 25-6pi r$.
$endgroup$
– KM101
Jan 7 at 22:55
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
I've attempted it again, in which I got r as root(25/3), is that correct? Edit : Surely, the 25π becomes nothing in the second derivative?
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:42
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Yes, you got the correct value of $r$. It may be better to rewrite $sqrt{frac{25}{3}}$ as $frac{5}{sqrt{3}} = frac{5sqrt{3}}{3}$. Now, $$f’’left(frac{5sqrt{3}}{3}right) = 25-6pileft(frac{5sqrt{3}}{3}right) = 25-10pisqrt{3} < 0$$ and $f’’(x) < 0$ implies the graph is concave down, hence proving the point is a maximum. From here, all that is left is finding $h$ now that you found $r$ and calculate the volume.
$endgroup$
– KM101
Jan 8 at 18:53
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
$begingroup$
Thank you very much! Also, are there any resources to use that help you develop these skills, if so, could you link them to me? It's just that worded questions always get me muddled (hence why I hate statistics). Have a nice day!
$endgroup$
– Jia Xuan Ng
Jan 8 at 18:58
|
show 1 more comment
$begingroup$
Hint:
The volume of a Cylinder $V=r^2pi h$
Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$
Now look for the maximum of that Function
answered Jan 7 at 20:42
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
Hint:
The volume of a Cylinder $V=r^2pi h$
Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$
Now look for the maximum of that Function
answered Jan 7 at 20:42
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
Hint:
The volume of a Cylinder $V=r^2pi h$
Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$
Now look for the maximum of that Function
answered Jan 7 at 20:42
A. PA. P
1085
$endgroup$
Hint:
The volume of a Cylinder $V=r^2pi h$
Using your formular for h we get $V=r^2pi*(frac{(A/2)}{r}-r)= pi((A/2)r-r^3)$
Now look for the maximum of that Function
answered Jan 7 at 20:42
A. PA. P
1085
answered Jan 7 at 20:42
A. PA. P
1085
answered Jan 7 at 20:42
A. PA. P
1085
answered Jan 7 at 20:42
A. PA. P
1085
1085
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$begingroup$
Welcome to StackExchange, Jia! A few comments: 1) Not everyone on here will be familiar with AS-levels, the site's users are worldwide 2) 'subbed back' - substituted, please! As for your question: 'This can then be [substituted] back into the formula to find r, but this is where I am completely muddled'. Why are you muddled here? That seems the correct way to go, and should leave a quadratic equation in $r$ for you to solve
$endgroup$
– bounceback
Jan 7 at 20:36