# How do you find the dimensions of a rectangular box that has the largest volume and surface area of 56 square units?

Mar 17, 2015

The box should be a cube and each dimension should be $\setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$ units. The maximum volume is $\setminus \frac{56 \setminus \sqrt{21}}{9} \setminus \approx 28.5$ cubic units.

Let $x$, $y$, and $z$ be the dimensions of the box. Then $V = x y z$ and $S = 2 x y + 2 x z + 2 y z$ are the volume and surface areas of the box.

It might make some intuitive sense to you that the answer is a cube so that $x = y = z$. If it does, then the surface area being 56 square units implies that $S = 6 {x}^{2} = 56$ so that ${x}^{2} = {y}^{2} = {z}^{2} = \setminus \frac{56}{6} = \setminus \frac{28}{3}$ and $x = y = z = \setminus \sqrt{\setminus \frac{28}{3}} = \setminus \frac{2 \setminus \sqrt{7}}{\setminus \sqrt{3}} = \setminus \frac{2 \setminus \sqrt{21}}{3}$ and $V = x y z = {\left(\setminus \frac{2 \setminus \sqrt{21}}{3}\right)}^{3} = \setminus \frac{8 \setminus \cdot 21 \setminus \sqrt{21}}{27} = \setminus \frac{56 \setminus \sqrt{21}}{9}$

More rigorously, we can use calculus:

Since $S = 56$, we can say that $2 z \left(x + y\right) = 56 - 2 x y$ so that $z = \setminus \frac{28 - x y}{x + y}$. This means we can write the volume as a function of two variables:

$V = \setminus \frac{x y \left(28 - x y\right)}{x + y} = \setminus \frac{28 x y - {x}^{2} {y}^{2}}{x + y} .$

The partial derivatives of this function are:

$\setminus \frac{\setminus \partial V}{\setminus \partial x} = \setminus \frac{\left(x + y\right) \left(28 y - 2 x {y}^{2}\right) - \left(28 x y - {x}^{2} {y}^{2}\right) \setminus \cdot 1}{{\left(x + y\right)}^{2}} = \setminus \frac{{y}^{2} \left(28 - {x}^{2} - 2 x y\right)}{{\left(x + y\right)}^{2}}$

and

$\setminus \frac{\setminus \partial V}{\setminus \partial y} = \setminus \frac{\left(x + y\right) \left(28 x - 2 {x}^{2} y\right) - \left(28 x y - {x}^{2} {y}^{2}\right) \setminus \cdot 1}{{\left(x + y\right)}^{2}} = \setminus \frac{{x}^{2} \left(28 - {y}^{2} - 2 x y\right)}{{\left(x + y\right)}^{2}} .$

Setting these both equal to zero results in the system of equations ${x}^{2} + 2 x y = 28 , {y}^{2} + 2 x y = 28$. The first of these equations can be solved for $y$ to get $y = \setminus \frac{28 - {x}^{2}}{2 x}$.

Plugging this into the second equation leads to ${\left(\setminus \frac{28 - {x}^{2}}{2 x}\right)}^{2} + 2 x \setminus \cdot \setminus \frac{28 - {x}^{2}}{2 x} = 28$. Multiplying both sides of this last equation by $4 {x}^{2}$ gives ${\left(28 - {x}^{2}\right)}^{2} + 112 {x}^{2} - 4 {x}^{4} = 112 {x}^{2}$, which reduces to $3 {x}^{4} + 56 {x}^{2} - 784 = 0$.

Using the quadratic formula, ${x}^{2} = \setminus \frac{- 56 \setminus \pm \setminus \sqrt{{56}^{2} - 4 \setminus \cdot 3 \setminus \cdot \left(- 784\right)}}{2} = \setminus \frac{- 56 \setminus \pm \setminus \sqrt{12544}}{6} = \setminus \frac{- 56 \setminus \pm 112}{6}$. The positive one is ${x}^{2} = \setminus \frac{56}{6} = \setminus \frac{28}{3}$ so that $x = \setminus \sqrt{\setminus \frac{28}{3}} = \setminus \frac{2 \setminus \sqrt{7}}{\setminus \sqrt{3}} = \setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$. The symmetry in the original equations implies that $y = \setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$ as well. If you use the equation $z = \setminus \frac{28 - x y}{x + y}$ you'll get $z = \setminus \frac{2 \setminus \sqrt{21}}{3} \setminus \approx 3.05$ as well. That this gives a maximum value can be checked with the 2nd derivative test.