Optimization problem (rectangle inscribed below parabola) - possible textbook mistake
$begingroup$
I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.
The problem:
A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.
c) What is that area?
Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.
If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.
Thank you.
calculus optimization
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
$endgroup$
add a comment |
$begingroup$
I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.
The problem:
A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.
c) What is that area?
Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.
If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.
Thank you.
calculus optimization
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
$endgroup$
1
$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19
add a comment |
$begingroup$
I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.
The problem:
A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.
c) What is that area?
Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.
If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.
Thank you.
calculus optimization
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
$endgroup$
I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.
The problem:
A rectangle is located below a parabola, which is given by $$y = 3x- frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.
c) What is that area?
Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.
If that's helpful, the textbook's answer for b) and c) are $3-sqrt{3} times 3$ and $9 + 9sqrt{3}$ respectively.
Thank you.
calculus optimization
calculus optimization
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
edited Jan 6 at 8:11
bru1987
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
asked Jan 6 at 8:00
bru1987bru1987
1,0691023
1,0691023
1
$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19
add a comment |
1
$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19
1
1
$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19
$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You made a slight error when trying to find the base:
$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$
You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.
Addition: From here, using $h = 3$ as you found, you get
$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$
Try plugging in $c = 3-sqrt{3}$ in $f(c)$:
$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum.
answered Jan 6 at 8:38
KM101KM101
5,9251524
$endgroup$
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
1
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
add a comment |
Your Answer
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1 Answer
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$begingroup$
You made a slight error when trying to find the base:
$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$
You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.
Addition: From here, using $h = 3$ as you found, you get
$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$
Try plugging in $c = 3-sqrt{3}$ in $f(c)$:
$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum.
answered Jan 6 at 8:38
KM101KM101
5,9251524
$endgroup$
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
1
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
add a comment |
$begingroup$
You made a slight error when trying to find the base:
$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$
You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.
Addition: From here, using $h = 3$ as you found, you get
$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$
Try plugging in $c = 3-sqrt{3}$ in $f(c)$:
$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum.
answered Jan 6 at 8:38
KM101KM101
5,9251524
$endgroup$
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
1
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
add a comment |
$begingroup$
You made a slight error when trying to find the base:
$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$
You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.
Addition: From here, using $h = 3$ as you found, you get
$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$
Try plugging in $c = 3-sqrt{3}$ in $f(c)$:
$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum.
answered Jan 6 at 8:38
KM101KM101
5,9251524
$endgroup$
You made a slight error when trying to find the base:
$$c = 3-sqrt{3} implies b = 2(3-c) = color{blue}{2left[3-left(3-sqrt{3}right)right]} = 2sqrt{3}$$
You forgot the $3$ in $(color{blue}{3}-c)$ and found $b = 2c$ instead.
Addition: From here, using $h = 3$ as you found, you get
$$A = bh iff A = 3left(2sqrt{3}right) = 6sqrt{3}$$
Try plugging in $c = 3-sqrt{3}$ in $f(c)$:
$$fleft(3-sqrt{3}right) = left(3-sqrt{3}right)^3-9left(3-sqrt{3}right)^2+18left(3-sqrt{3}right) = 6sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum.
answered Jan 6 at 8:38
KM101KM101
5,9251524
edited Jan 6 at 9:36
answered Jan 6 at 8:38
KM101KM101
5,9251524
answered Jan 6 at 8:38
KM101KM101
5,9251524
answered Jan 6 at 8:38
KM101KM101
5,9251524
5,9251524
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
1
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
add a comment |
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
1
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
thank you I've fixed that mistake. Now I have $2 sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 sqrt{3} m^2$. Am I good? Thanks for your help!
$endgroup$
– bru1987
Jan 6 at 9:16
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
Yes, that’s the correct answer!
$endgroup$
– KM101
Jan 6 at 9:24
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
$begingroup$
thank you and have a great day =)
$endgroup$
– bru1987
Jan 6 at 9:26
1
1
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
$begingroup$
Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue.
$endgroup$
– KM101
Jan 6 at 9:31
add a comment |
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$begingroup$
$base=2sqrt{3}$
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:19