Find the dimensions of a rectangle of area $400 \mathrm { ft } ^ { 2 }$ that has the smallest possible perimeter.
A) $4 \mathrm { ft }$ by $100 \mathrm { ft }$
B) $1 \mathrm { ft }$ by $400 \mathrm { ft }$
C) $2 \mathrm { ft }$ by $200 \mathrm { ft }$
D) $20 \mathrm { ft }$ by $20 \mathrm { ft }$

Question

Quizplus · Accepted Answer

Let the length be $l$ and the width be $w$. We want to minimize the perimeter $P=2l+2w$, subject to the constraint $lw=400$. We can solve for $w$ in terms of $l$ from the equation $lw=400$, to get $w=400/l$. We can now substitute this expression for $w$ into the equation for $P$, to get $P=2l+2(400/l)$.
We can now apply calculus to find the minimum value of $P$. Taking the derivative of $P$ with respect to $l$ and setting it equal to zero, we get:
$$2-800/l^2=0$$
Solving for $l$, we get $l=20$, and therefore $w=20$, which gives us a rectangle with dimensions $20\mathrm{ft}	imes 20\mathrm{ft}$, and a perimeter of $P=2(20)+2(20)=\boxed{40	ext{ ft}}$. We could also check that this is a minimum by taking the second derivative of $P$ with respect to $l$ and verifying that it is positive.

Find the Dimensions of a Rectangle of Area 400ft2400 \mathrm { ft } ^ { 2 }400ft2

Find the Dimensions of a Rectangle of Area $400 \mathrm { ft } ^ { 2 }$