Question

Find the dimensions of a rectangle with area 216 m2 whose perimeter is as small as possible. (if both values are the same number, enter it into both blanks.)

Answer 1

1. Let the width of the rectangle be a, and its height be b.

a*b=216, so b=216/a

2. the perimeter of the rectangle is P=2a+2b=2a+2(216/a)=2a+432/a

since a varies, and P also varies accordingly, we can write P as a function in variable a as follows:

$P(a)=2a+432a ^{-1}$

3. Taking the derivative we find the critical points where we have the largest or smallest values the function may take:

$P'(a)=2-432a ^{-2}=2- \frac{432}{ a^{2} }$

$2- \frac{432}{ a^{2} }=0$

$\frac{432}{ a^{2} }=2$

$\frac{216}{ a^{2} }=1$

$a= \sqrt{216}=14.7$ or $a= \sqrt{216}=-14.7$

4. Since $P(a)=2a+432a ^{-1}=2a+ \frac{432}{a}$ would be negative with the negative root, the largest value of P is when a=14.7


5. b=216/a=216/14.7=14.7


Answer: [14.7, 14.7]   (or [square root 216, square root 216])