Optimization of rectangle. Find size of fence

Question

I've got this, and im unsure how to proceed. I need to build a fence. There is a building next to the fence, which is why i only need $3$ sides of the fence and it shall be a rectangle. So this applies: $o=2x+y$ The area shall be $558m^2$ So $558=l*w$ How can i find the smallest size of the fences sides?

I've tried the following: 558=x*y y=558-x Then put it in this: o = 2x+y So: 2x+(558-x) -> -x^2+558x Then differentiated it to: 554-4x Set it to 0 and found x to be 139.5

unfortunately that is not correct

Please consider including your thoughts. There are multiple ways to solve this problem. Have you learned about derivatives yet? — Sina Babaei Zadeh, Nov 08 '19 at 00:15
I Can't see how it relates. He got the width. I've got the area on 558m2. — Exobit, Nov 08 '19 at 00:34
This is probably the oldest exercise in the history of Calculus textbooks, and it is certain that (something equivalent to) it has been asked and answered here many, many times. I encourage you to search the site for previous appearances. — Gerry Myerson, Nov 08 '19 at 00:50
I tried. I've found alot of similarities, and they didn't work in my case — Exobit, Nov 08 '19 at 00:58
Show us one of the similarities, and tell us why it doesn't work, please. — Gerry Myerson, Nov 08 '19 at 11:36
I'm voting to close this question as off-topic because OP has abandoned it. — Gerry Myerson, Nov 10 '19 at 11:33

Narasimham · Answer 1 · 2019-11-08T10:56:14.023

This a standard maxima/minima problem. Shall be brief. Let Length and Width be $(x,y)$ respectively. $$ \sigma= 2 x +y ,\quad A= xy, \quad y= \frac{A}{x}... \sigma= 2x+\frac{A}{x}$$ Differentiate, equate to zero, solve for $(x_{opt},\quad y_{opt} ) $ and finally find $\sigma_{min}$ $$\frac{d\sigma}{dx}=2-\frac{A}{x^2}=0 \rightarrow (x_{opt}=\sqrt{A/2}, y_{opt}=\sqrt{2 A}) $$

$$\rightarrow \sigma_{min}=2\sqrt{A/2}+\sqrt{2 A}=2\sqrt{2} A= 2\sqrt{ 2} * 558.0 \approx 1578.262 $$

Note that

$$ {x_{opt}} =\sqrt{558/2}= 16.708,{y_{opt}}=\sqrt{558*2} =33.4066 ;$$ $$ \frac{x_{opt}}{y_{opt}}= \dfrac{\sqrt{A/2}} {\sqrt{2 A}} =\dfrac12 $$

In your work it should be $y=558/x, $ but not $ y=558- x.$

score 0 · Answer 2 · answered Nov 08 '19 at 07:31

Let $l$ and $w$ represent the length and width of the rectangle. We are told that $l=358w^{-1}$. We want to find values that minimize $2w+l$. So $$f(w)=2w+558w^{-1}$$ Differentiating and setting the derivative equal to zero gives us $$f'(w)=2-558w^{-2}=0\\2=558w^{-2}\\2w^2=558\\w=\pm\sqrt{279}=\pm3\sqrt{31}$$

Optimization of rectangle. Find size of fence

2 Answers2