# Thread: Optimization problem, perimeter of a rectangle

1. ## Optimization problem, perimeter of a rectangle

Dear forum members,

A problem asks me to show that a rectangle of constant area A has a minimal perimeter when the rectangle is a square.

I make the equation for the area and differentiate it with respect to x.

$P’(x)= 2-\frac{2A}{x^2}$

Solving for the zero points

$2x^2-2A=0$

At this stage, am I allowed to substitute A=xh into the equation and solve it as
$2x^2-2xh=0$

$2x(x-h)=0$
x=0 or x=h

or should I solve it as

$x=-\sqrt{A}$ or $x=\sqrt{A}$

because evidently the answer is different .

2. Originally Posted by Coach
...show that a rectangle of constant area A has a minimal perimeter when the rectangle is a square.
I'm not sure what your initial equations were or what you were doing with them...?

Try starting with the "area" and "perimeter" formulas for a rectangle with height "h" and width "w".

. . . . . $A\, =\, hw$

. . . . . $P\, =\, 2h\, +\, 2w$

Solving the "area" formula for one of the dimensions (keeping in mind that the area, in this case, is a constant value), you get:

. . . . . $h\, =\, \frac{A}{w}$

Subsitituting, you get:

. . . . . $P(w)\, =\, \frac{2A}{w}\, +\, 2w$

Differentiating, and keeping in mind that dA/dw = 0, you get:

. . . . . $P'(w)\, =\, \frac{2\frac{dA}{dw}w\, -\, 2A}{w^2}\, +\, 2\, =\, -\frac{2A}{w^2}\, +\, 2$

Setting this equal to zero, you get:

. . . . . $-\frac{2A}{w^2}\, +\, 2\, =\, 0$

. . . . . $-2A\, +\, 2w^2\, =\, 0$

. . . . . $w^2\, =\, A$

. . . . . $w\, =\, \sqrt{A}$

(Since the width cannot be negative, you can ignore the negative square root.)

Since:

. . . . . $A\, =\, hw$

...and since the value you want is at w = sqrt[A], then:

. . . . . $A\, =\, h\sqrt{A}$

...then:

. . . . . $h\, =\, \frac{A}{\sqrt{A}}\, =\, \sqrt{A}$

What sort of rectangle has the same width as height?