Question

Perimeter of a rectangle is 500 cm. How do you find the maximum area of the rectangle? What is the length and the width?

Answer 1

$$
P=2x+2y=5m\\
A=x\times y\\
x>0\\
y>0
$$
We need to maximize A subject to P.
$$
2y = 5-2x\\
y=\cfrac{5}{2}-x\\
A=x\times y=x\times (\cfrac{5}{2}-x)=\cfrac{5}{2}x-x^2\\
$$
Where \(P \) is Perimiter and \(A\) is Area.

To maximize A, take derivative and set equal to zero:
$$
\cfrac{d A}{dx}=\cfrac{d}{dx}\left(\cfrac{5}{2}x-x^2 \right)=\cfrac{5}{2}-2x=0\\
\implies 2x=\cfrac{5}{2}\implies x=\cfrac{5}{4}\\
\implies A=\cfrac{5}{2}\cfrac{5}{4}-\left(\cfrac{5}{4}\right )^2\\
\qquad \qquad = \cfrac{25}{8}-\cfrac{25}{16}=\cfrac{25\times 2-25}{16}=\cfrac{25}{16}
$$
Therefore, the maximum area is \(\cfrac{25}{16}~m^2\).