how to find the dimensions of the rectangular box having the largest volume and surface area 150 square units?
I was afraid you'd say that.
The max volume for a closed box with a fixed surface area will be a cube.
Here is the work ...
let x , y , and z be the length, width, and height.
$\displaystyle V = xyz$
$\displaystyle A = 150 = 2xy + 2xz + 2yz$
$\displaystyle xy + xz + yz = 75$
$\displaystyle z = \frac{75-xy}{x+y}$
$\displaystyle V = \frac{75xy - x^2y^2}{x+y}$
$\displaystyle \frac{\partial V}{\partial x} = \frac{(x+y)(75y - 2xy^2)-(75xy - x^2y^2)(1)}{(x+y)^2}$
$\displaystyle \frac{\partial V}{\partial x} = \frac{75xy - 2x^2y^2 -2xy^3 + 75y^2 - 75xy + x^2y^2}{(x+y)^2}$
$\displaystyle \frac{\partial V}{\partial x} = \frac{75y^2-x^2y^2-2xy^3}{(x+y)^2}$
since $\displaystyle y \neq 0$
$\displaystyle \frac{\partial V}{\partial x} = \frac{75-2xy-x^2}{(x+y)^2}$
$\displaystyle \frac{\partial V}{\partial y} = \frac{(x+y)(75x - 2x^2y)-(75xy - x^2y^2)(1)}{(x+y)^2}$
$\displaystyle \frac{\partial V}{\partial y} = \frac{75x^2 - 2x^3y + 75xy - 2x^2y^2 - 75xy + x^2y^2}{(x+y)^2}$
$\displaystyle \frac{\partial V}{\partial y} = \frac{75x^2 - 2x^3y - x^2y^2}{(x+y)^2}$
since $\displaystyle x \neq 0$ ...
$\displaystyle \frac{\partial V}{\partial y} = \frac{75 - 2xy - y^2}{(x+y)^2}$
set both partials = 0 ...
$\displaystyle 75-2xy-x^2 = 0$
$\displaystyle 75 - 2xy - y^2 = 0$
combine both equations ...
$\displaystyle x^2 - y^2 = 0$
$\displaystyle (x-y)(x+y) = 0$
since $\displaystyle x \neq -y$ ... $\displaystyle x = y$
$\displaystyle z = \frac{75-xy}{x+y} = \frac{75-x^2}{2x}$
$\displaystyle V = x^2z = \frac{75x - x^3}{2}$
$\displaystyle \frac{dV}{dx} = \frac{75 - 3x^2}{2} = 0$
$\displaystyle x = 5$
$\displaystyle y = 5$
$\displaystyle z = 5$
the "cube"