Show that the largest rectangle with a perimeter of $20$ meters is a square.

Question

Show that the largest rectangle with a perimeter of $20$ meters is a square.

Answer 1

Hint: You know that $2W+2H=20$. Use that to compute the maximum of $W\cdot H$.

Answer 2

Perimeter $P=2h+2w=20$. So $h+w=10 \implies h=10-w$

$hw=w(10-w)=10w-w^2$

Find the maximum of this function by setting the derivative equal to 0 and solving for $w$.

Answer 3

To expand on the other answers, I posted a solution below that is hidden. Simply hover your mouse over it to check your work.

As Andrew noted, the perimeter is given by $2h+2w = 20$ where $h$ is the height and $w$ is the width of the rectangle. Dividing through by $2$ gives us $h + w = 10$. Rewriting $h + w = 10$ as $w = 10 - h$, we can plug that expression for $w$ into our area formula. So, we are left with $h\cdot{w} = h(10-h) = 10h-h^2$. Now, since we are trying to maximize this, we want to take the derivative and set it equal to $0$ (finding critical points - an essential tool in introductory calculus courses.) So, the derivative would be $10 - 2h$. Setting that equal to $0$ and solving for $h$ leaves us $h = 5$. Plugging in $h = 5$ into our expression $h + w = 10$ tells us that the width must also be $5$ meters. So, since the height and width are both $5$ meters, the largest rectangle with a perimeter of $20$ meters is indeed a square.

Answer 4

And you can avoid taking the derivative by writing $$10 w - w^2 = 25 - 25 + 10 w - w^2 = 25 - (25 -10w + w^2) = 25 - (5-w)^2 \le 25 $$ with equality if and only if $w = 5$.

This also works for any value, not just $20$.

Answer 5

The short and long sides of the rectangle have lengths $5-x$ and $5+x$ respectively, for some $x$ between $0$ and $5.$ The area is $A=(5-x)(5+x)=25-x^2.$ So how should you pick $x$ to make the area the largest?