Question

a rancher must design a rectangular corral with an area of 400 square meters. she decides to make a corral that measures 10 meters by 40 meters (which has the correct total area). how much fence is needed for this corral? has the rancher found the most economical solution? can you find another design for the corral that requires less fence but still provides 400 square meters of corral? the rancher needs meters of fence.

Explanation:

Step1: Recall the perimeter formula for a rectangle

The perimeter of a rectangle is $P = 2l+2w$ and the area is $A=lw$, where $l$ is the length and $w$ is the width. Given $A = lw=400$. So, $l=\frac{400}{w}$.

Step2: Substitute $l$ into the perimeter formula

$P = 2(\frac{400}{w})+2w=\frac{800}{w}+2w$.

Step3: Take the derivative of the perimeter function

Let $y=\frac{800}{w}+2w = 800w^{- 1}+2w$. The derivative $y^\prime=-800w^{-2}+2=\frac{-800 + 2w^{2}}{w^{2}}$.

Step4: Set the derivative equal to zero to find the critical - points

$\frac{-800 + 2w^{2}}{w^{2}}=0$. Then $-800 + 2w^{2}=0$. So, $2w^{2}=800$, $w^{2}=400$, and $w = 20$ (we consider the positive value since width cannot be negative).

Step5: Find the length

If $w = 20$, then $l=\frac{400}{20}=20$. The perimeter of a square (a special rectangle) with side length $s = 20$ is $P=4s=80$ meters. The perimeter of the $10\times40$ rectangle is $P=2(10 + 40)=100$ meters.

Answer:

80