How can I optimize the are of a rectangle inscribed between a line and the x axis?
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I have these two images. For the first one, I have a parabola and the rectangle inside it. It says that the function on the right is the area function of the rectangle.
I understand how to do the first optimization problem. I have
The point where the square touches the parabola is (x,y).
The area is 2xy
If i consider only the positive x values
Then the width is y or $y=9-x^2$
So substituting it into the area function we have
$A = 2x(9-x^2) = 18x-2x^3$
Now to optimize, I would take the derivative of the area:
$18-6x^2$
$x=pm sqrt{3}$
Then I plug back in my positive x value to get $y = 6$
This the $A = 2(sqrt{3})6 = 12sqrt{3}$ and the length is $2sqrt{3}$ and the width is 6. Is this correct?
For the second problem, I don't know how to approach it.
The line is $y = 6-2x$
Then similarly the point where the rectangle touches the line is $(x,y)$
The area is just $xy$ this time
$y= 6-2x$ and so the area is $x(6-2x) = 6x-2x^2$
Then to optimize I would take the derivative and get
$6-4x$ so $x = 1.5$
Then $y = 6-2(1.5) = 3 $
So the area = $3(1.5) = 4.5$
Is this correct? I'm confused because the graphs shown in the question show those same dimensions so I'm not sure if I'm finding the largest dimensions of the rectange?
algebra-precalculus
asked Jan 14 at 20:30
user130306user130306
45319
$endgroup$
add a comment |
$begingroup$
I have these two images. For the first one, I have a parabola and the rectangle inside it. It says that the function on the right is the area function of the rectangle.
I understand how to do the first optimization problem. I have
The point where the square touches the parabola is (x,y).
The area is 2xy
If i consider only the positive x values
Then the width is y or $y=9-x^2$
So substituting it into the area function we have
$A = 2x(9-x^2) = 18x-2x^3$
Now to optimize, I would take the derivative of the area:
$18-6x^2$
$x=pm sqrt{3}$
Then I plug back in my positive x value to get $y = 6$
This the $A = 2(sqrt{3})6 = 12sqrt{3}$ and the length is $2sqrt{3}$ and the width is 6. Is this correct?
For the second problem, I don't know how to approach it.
The line is $y = 6-2x$
Then similarly the point where the rectangle touches the line is $(x,y)$
The area is just $xy$ this time
$y= 6-2x$ and so the area is $x(6-2x) = 6x-2x^2$
Then to optimize I would take the derivative and get
$6-4x$ so $x = 1.5$
Then $y = 6-2(1.5) = 3 $
So the area = $3(1.5) = 4.5$
Is this correct? I'm confused because the graphs shown in the question show those same dimensions so I'm not sure if I'm finding the largest dimensions of the rectange?
algebra-precalculus
asked Jan 14 at 20:30
user130306user130306
45319
$endgroup$
$begingroup$
What is necessary in the statement of the second problem for it to have a solution?
$endgroup$
– William Elliot
Jan 14 at 21:54
$begingroup$
it has to touch the line?
$endgroup$
– user130306
Jan 14 at 21:59
$begingroup$
That it already does. Why is the area xy?
$endgroup$
– William Elliot
Jan 15 at 3:44
add a comment |
$begingroup$
I have these two images. For the first one, I have a parabola and the rectangle inside it. It says that the function on the right is the area function of the rectangle.
I understand how to do the first optimization problem. I have
The point where the square touches the parabola is (x,y).
The area is 2xy
If i consider only the positive x values
Then the width is y or $y=9-x^2$
So substituting it into the area function we have
$A = 2x(9-x^2) = 18x-2x^3$
Now to optimize, I would take the derivative of the area:
$18-6x^2$
$x=pm sqrt{3}$
Then I plug back in my positive x value to get $y = 6$
This the $A = 2(sqrt{3})6 = 12sqrt{3}$ and the length is $2sqrt{3}$ and the width is 6. Is this correct?
For the second problem, I don't know how to approach it.
The line is $y = 6-2x$
Then similarly the point where the rectangle touches the line is $(x,y)$
The area is just $xy$ this time
$y= 6-2x$ and so the area is $x(6-2x) = 6x-2x^2$
Then to optimize I would take the derivative and get
$6-4x$ so $x = 1.5$
Then $y = 6-2(1.5) = 3 $
So the area = $3(1.5) = 4.5$
Is this correct? I'm confused because the graphs shown in the question show those same dimensions so I'm not sure if I'm finding the largest dimensions of the rectange?
algebra-precalculus
asked Jan 14 at 20:30
user130306user130306
45319
$endgroup$
I have these two images. For the first one, I have a parabola and the rectangle inside it. It says that the function on the right is the area function of the rectangle.
I understand how to do the first optimization problem. I have
The point where the square touches the parabola is (x,y).
The area is 2xy
If i consider only the positive x values
Then the width is y or $y=9-x^2$
So substituting it into the area function we have
$A = 2x(9-x^2) = 18x-2x^3$
Now to optimize, I would take the derivative of the area:
$18-6x^2$
$x=pm sqrt{3}$
Then I plug back in my positive x value to get $y = 6$
This the $A = 2(sqrt{3})6 = 12sqrt{3}$ and the length is $2sqrt{3}$ and the width is 6. Is this correct?
For the second problem, I don't know how to approach it.
The line is $y = 6-2x$
Then similarly the point where the rectangle touches the line is $(x,y)$
The area is just $xy$ this time
$y= 6-2x$ and so the area is $x(6-2x) = 6x-2x^2$
Then to optimize I would take the derivative and get
$6-4x$ so $x = 1.5$
Then $y = 6-2(1.5) = 3 $
So the area = $3(1.5) = 4.5$
Is this correct? I'm confused because the graphs shown in the question show those same dimensions so I'm not sure if I'm finding the largest dimensions of the rectange?
algebra-precalculus
algebra-precalculus
asked Jan 14 at 20:30
user130306user130306
45319
asked Jan 14 at 20:30
user130306user130306
45319
asked Jan 14 at 20:30
user130306user130306
45319
asked Jan 14 at 20:30
user130306user130306
45319
asked Jan 14 at 20:30
user130306user130306
45319
45319
$begingroup$
What is necessary in the statement of the second problem for it to have a solution?
$endgroup$
– William Elliot
Jan 14 at 21:54
$begingroup$
it has to touch the line?
$endgroup$
– user130306
Jan 14 at 21:59
$begingroup$
That it already does. Why is the area xy?
$endgroup$
– William Elliot
Jan 15 at 3:44
add a comment |
$begingroup$
What is necessary in the statement of the second problem for it to have a solution?
$endgroup$
– William Elliot
Jan 14 at 21:54
$begingroup$
it has to touch the line?
$endgroup$
– user130306
Jan 14 at 21:59
$begingroup$
That it already does. Why is the area xy?
$endgroup$
– William Elliot
Jan 15 at 3:44
$begingroup$
What is necessary in the statement of the second problem for it to have a solution?
$endgroup$
– William Elliot
Jan 14 at 21:54
$begingroup$
What is necessary in the statement of the second problem for it to have a solution?
$endgroup$
– William Elliot
Jan 14 at 21:54
$begingroup$
it has to touch the line?
$endgroup$
– user130306
Jan 14 at 21:59
$begingroup$
it has to touch the line?
$endgroup$
– user130306
Jan 14 at 21:59
$begingroup$
That it already does. Why is the area xy?
$endgroup$
– William Elliot
Jan 15 at 3:44
$begingroup$
That it already does. Why is the area xy?
$endgroup$
– William Elliot
Jan 15 at 3:44
add a comment |
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$begingroup$
What is necessary in the statement of the second problem for it to have a solution?
$endgroup$
– William Elliot
Jan 14 at 21:54
$begingroup$
it has to touch the line?
$endgroup$
– user130306
Jan 14 at 21:59
$begingroup$
That it already does. Why is the area xy?
$endgroup$
– William Elliot
Jan 15 at 3:44