perimeter of the regtangle

Question

if the area of rectangle is 40, which of the following could be the perimeter of the rectangle? Indicate all such areas.

A) 20

B) 40

C) 200

D)400

E)2000

F)4000

Answers are B,C,D,E,F

Answer 1

Suppose the length of the rectangle is $x$ and the width of the rectangle is $y$.Write your given information in terms of algebra:

$$xy=40$$

$$x+x+y+y=2x+2y=P$$

Of course we have:

$$x > 0, y > 0$$

Now use the first equation to solve for $y$ and write the equation of the perimeter in terms of only $x$.

Minimize and maximize the perimeter using standard methods $f'(x)=0$ and noting the end behaviors.

$$y=\frac{40}{x}$$. $$P=2x+\frac{80}{x}$$. $$x \to 0^+ \implies P \to \infty, x \to \infty \implies P \to \infty$$. Now let's solve for the minimum: $$f'(x)=0 \implies 2-\frac{80}{x^2}=0 \implies x=\pm \sqrt{40}=\pm 2\sqrt{10}$$. $$x > 0 \implies x=2\sqrt{10} \implies \text{min}(P)=4\sqrt{10}+\frac{80}{2\sqrt{10}}=8\sqrt{10}$$ considering our already evaluated end behaviors on our domain $x \in (0,\infty)$. Above I multiplied by the fraction $\frac{80}{2\sqrt{10}}$ by $\frac{\sqrt{10}}{\sqrt{10}}=1$ to take the square root out of the denominator. Hence our only constraint on the perimeter is $P \geq 8\sqrt{10}$. Without a calculator you can figure $8(4)>8\sqrt{10}>8(3)$ from which you can eliminate the answer you need to eliminate and keep the answers you need to keep.