Optimization problem (Max area of a combined semi circle and a square)

[Mathman2013]

Homework Statement

Optimization problem (Max area of a combined semi circle and a square)

Relevant Equations

f (w) = (4 - 9/7*w)*w + 11/28*w^2

A figure is made from a semi circle and square. With the following dimensions, width = w, and length = l.

Find the maximum area when the combined perimiter is 8 meter.

I first try to construct the a function for the perimeter.

2*l + w + 22/7*w/2 = 8 - > l = 4 - (9*w)/7

Next I insert this into a function which is suppose to define the area.

f(w) = l*w + 1/2*22/7*(w/2)^2 and substitute l with l = 4 - (9*w)/7

which results in f(w) = (4 - 9/7*w)*w + 11/28*w^2

I find the derivative of this function f'(w) = -(25*w)/14 + 4 and solve f'(w) = 0 and find that w is suppose to be 2.24 m

However the solution manual says w = 1.86 m.

So what am I doing wrong?

 

[BvU]

You forget to make a sketch. What is l and what is w for a square ? and for a circle ?

 

[Mathman2013]

You forget to make a sketch. What is l and what is w for a square ? and for a circle ?

Thank you for your reply,

The drawing in the textbook look like this. Any thoughts on what I am doing wrong? In regards to my calculations?

 

[Delta2]

I can't find what you doing wrong, wolfram also says that w=56/25. My only doubt is regarding the scheme, is that an exact image from the problem at the book?

 

[Mathman2013]

Thank your answer, the image in the textbook in the included file. The text is in danish, but simply says: A flowerbed is comprised of a semi-circle and square. The perimeter of the entire must be 8 meters in total.

How long does the sides of the square part have to be? To get the maximum area of the square part of flowerbed?

And I constructed a function for the perimeter in initial post post. Calling the length l and width w. Next isolated with respect to l, and next found the area function. f(w) which calculated the derivative of. and solved the equation f'(w) = 0 and found w and used this result to find L.

Can it be an error in the problem? Or an error in the solution manual?

 

[Delta2]

unfortunately I don't know Danish to read for myself, but does it say "the maximum area of the square part" of flowerbed?. IF so I think it asks to maximize the area of the rectangle with sides l and w, that is to maximize $l\cdot w$, ignoring the area of the circular part.

 

[Mathman2013]

Thank you again for your answer, no danish is a small language :)

I try again,
But if the function for the perimeter is 2*l+w + pi *w/2 = 8,(if we assume l is length of the square part, and w is the width) so that l = 4 - (9*w)/7

If I insert this into the function f(w) = w*( 4 - (9*w)/7) - > f(w) = 4*w- (9*w^2)/7

and solve via the derivative, then I get. solve(f'(w) = 0) , then w = 1.555555556 and l = 2.

Which should according to my math give me the optimal values for l and w.

Any thoughts on my calculations?

 

[Delta2]

I think you are correct but we still far from the solution of 1.86 given by the textbook...
I don't know what else to think of, sorry.

 

[HallsofIvy]

You confused things by saying "square". If w is not equal to l, that is not a square. Taking w to be the length of the top and bottom of the rectangle and l to be the length of the right and left sides, then the circle has radius w/2. The distance around the semicircle is \pi w/2 and the distance around the three sides of the rectangle is 2l+ w. So the total distance around the figure is \frac{\pi w}{2}+ 2l+ w= 8. The area of the semicircle is \frac{\pi w^2}{4} and the area of the rectangle is lw. The problem is asking you to maximize \frac{\pi w^2}{4}+ lw subject to the condition that \frac{\pi w}{2}+ 2l+ w= 8. The simplest way to do that is to solve \frac{\pi w}{2}+ 2l+ w= 8 for l so that you can write \frac{pi w^2}{4}+ lw as a function of w only.

 

[Mathman2013]

Dear HallsoftIvy,

If I try to solve in Maple, then I get that w is 4?

 

[BvU]

semi-circle and square.

Should be rectangle. 2.24 is correct, book answer wrong.

 

[BvU]

If I try to solve in Maple, then I get that w is 4?

Check the half-circle area expression...