Originally Posted by

**Jacobpm64** Of all rectangles with given area, $\displaystyle A$, 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

$\displaystyle A = xy$

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

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

$\displaystyle 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).

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

$\displaystyle x - \frac{A^2}{x^3} = 0$

$\displaystyle x^4 - A^2 = 0$

$\displaystyle x^4 = A^2$

$\displaystyle x = \sqrt{A}$

Plug it back in to the area to find the other dimension.

$\displaystyle A = \sqrt{A} * y$

$\displaystyle y = \frac{A}{\sqrt{A}}$

$\displaystyle y = \sqrt{A}$

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

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