# Math Help - Minimizing Diagonals of rectangles

1. ## Minimizing Diagonals of rectangles

Of all rectangles with given area, $A$, which has the shortest diagonals?

I'm not sure what this question is asking, but I tried. I'll show my work, Please tell me if I did it correctly or help me if I didn't.

x = length
y = width
A = area

$A = xy$

The diagonal would be $d = \sqrt{x^2 + y^2}$.

Because I want to minimize the diagonals, I solved for y in the area function and substituted it into the formula for the diagonal.

$d = \sqrt{x^2 + (\frac{A}{x})^2}$

I'll take the derivative of the diagonal and set it equal to zero to minimize it (here i treated A as a constant... I'm not sure if i can do this.. this may be my mistake.. I'm not sure).

$d' = \frac{1}{2} (x^2 + (\frac{A}{x})^2))^{-\frac{1}{2}}(2x - \frac{2A^2}{x^3}) = 0$
$x - \frac{A^2}{x^3} = 0$
$x^4 - A^2 = 0$
$x^4 = A^2$
$x = \sqrt{A}$

Plug it back in to the area to find the other dimension.
$A = \sqrt{A} * y$
$y = \frac{A}{\sqrt{A}}$
$y = \sqrt{A}$

The dimensions should be $\sqrt{A} \quad x\quad \sqrt{A}$.

Is this what they wanted? And is this correct? I have no clue =\

2. The answer is correct but I haven't looked through your work, notice that when a side is the square root of A then the rectangle is a square.

3. Originally Posted by Jacobpm64
Of all rectangles with given area, $A$, which has the shortest diagonals?

I'm not sure what this question is asking, but I tried. I'll show my work, Please tell me if I did it correctly or help me if I didn't.

x = length
y = width
A = area

$A = xy$

The diagonal would be $d = \sqrt{x^2 + y^2}$.

Because I want to minimize the diagonals, I solved for y in the area function and substituted it into the formula for the diagonal.

$d = \sqrt{x^2 + (\frac{A}{x})^2}$

I'll take the derivative of the diagonal and set it equal to zero to minimize it (here i treated A as a constant... I'm not sure if i can do this.. this may be my mistake.. I'm not sure).

$d' = \frac{1}{2} (x^2 + (\frac{A}{x})^2))^{-\frac{1}{2}}(2x - \frac{2A^2}{x^3}) = 0$
$x - \frac{A^2}{x^3} = 0$
$x^4 - A^2 = 0$
$x^4 = A^2$
$x = \sqrt{A}$

Plug it back in to the area to find the other dimension.
$A = \sqrt{A} * y$
$y = \frac{A}{\sqrt{A}}$
$y = \sqrt{A}$

The dimensions should be $\sqrt{A} \quad x\quad \sqrt{A}$.

Is this what they wanted? And is this correct? I have no clue =\
Yep, you got it right!

-Dan