Question

Find the dimensions of the box with volume 4096 cm3 that has minimal surface area. (Let x, y, and z be the dimensions of the box.)

Answer 1

The dimensions of the box with a volume of 4096 cm³ that has a minimal surface area are 16cm×16cm×16 cm.

A solid object's surface area is a measurement of the overall space that the object's surface takes up. It is also known as the overall area of a three-dimensional shape's surface. The formula to calculate surface area is S = 2(xy+yz+zx) where x is length, y is width and z is the height of the object.  

Given the volume is 4096 cm³.

Then, V = xyz = 4096. Deriving the value of z from this, we get z = 4096/xy. Substitute value of z in S = 2(xy+yz+zx), we get,

$\begin{aligned}S&=2\left(xy+y\times\frac{4096}{xy}+x\times\frac{4096}{xy}\right)\\&=2\left(xy+\frac{4096}{x}+\frac{4096}{y}\right)\\&=2xy+\frac{8192}{x}+\frac{8192}{y}\\&=2xy+8192\left(\frac{1}{x}+\frac{1}{y}\right)\end{aligned}$

Differentiating S with respect to x, we get,

$\frac{dS}{dx}=2y-\frac{8192}{x^2}$

Equating this to zero and multiplying with x² we get,

$\begin{aligned}2y-\frac{8192}{x^2}&=0\\2x^2y-8192&=0\\x^2y&=\frac{8192}{2}\\x^2y&=4096\end{aligned}$

Differentiating S with respect to y, we get,

$\frac{dS}{dy}=2x-\frac{8192}{y^2}$

Equating this to zero and multiplying with y² we get,

$\begin{aligned}2x-\frac{8192}{y^2}&=0\\2xy^2-8192&=0\\xy^2&=\frac{8192}{2}\\xy^2&=4096\end{aligned}$

Dividing x²y by xy², we get,

$\begin{aligned}\frac{x^2y}{xy^2}&=\frac{4096}{4096}\\\frac{x}{y}&=1\\x&=y\end{aligned}$

Substitute x = y in x²y = 4096, we get,

$\begin{aligned}y^3&=4096\\y&=16=x\end{aligned}$

Then, the z value is

$\begin{aligned}z &= \frac{4096}{16\times 16}\\&=16\end{aligned}$

The required answer is 16cm×16cm×16 cm.

To know more about the surface area:

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