Answer to Question #206099 in Calculus for Varun

Question

A unique package is made up of cube with cylinder on top. The diameter of the cylinder equals the length of the cube. If the total volume is 50 cubic cm. What dimensions of the package will minimize the surface area of the package?

Answer 1

Let $x=$ the length of the cube, $y=$ the height of the cylinder.

The volume of the cube is $V_{cube}=x^3.$

The volume of the cylinder is $V_{cylinder}=\pi(\dfrac{x}{2})^2y.$

The total volume is $50\ cm^3$

$x^3+\dfrac{\pi}{4}x^2y=50$

Solve for $y$

$y=\dfrac{200-4x^3}{\pi x^2}$

The surface area of the package is the sum of the surface area of the cube and the surface area of the cylinder

$A=6x^2+2\pi(\dfrac{x}{2})^2 +2\pi(\dfrac{x}{2})y$

$=6x^2+\dfrac{\pi}{2}x^2+\pi xy$

Substitute $y=\dfrac{200-4x^3}{\pi x^2}$

$A=A(x)=6x^2+\dfrac{\pi}{2}x^2+\dfrac{200-4x^3}{x}, x>0$

Find the first derivative with respect to $x$

$A'(x)=(6x^2+\dfrac{\pi}{2}x^2+\dfrac{200-4x^3}{x})'=$

$=12x+\pi x+\dfrac{-12x^3-(200-4x^3)}{x^2}$

$=\dfrac{12x^3+\pi x^3-12x^3-200+4x^3}{x^2}$

$=\dfrac{-200+(4+\pi)x^3}{x^2}$

Find the critical numbers

$A'(x)=0=>\dfrac{-200+(4+\pi)x^3}{x^2}=0$

$x=\sqrt[3]{\dfrac{200}{4+\pi}}$

If $0<x<\sqrt[3]{\dfrac{200}{4+\pi}}, A'(x)<0, A(x)$ decreases.

If $x>\sqrt[3]{\dfrac{200}{4+\pi}}, A'(x)>0, A(x)$ increases.

The function $A(x)$ has a local minimum at $x=\sqrt[3]{\dfrac{200}{4+\pi}}.$

Since the function $A(x)$ has the only extremum for $x>0,$ then the function $A(x)$ has the absolute minimum at $x=\sqrt[3]{\dfrac{200}{4+\pi}}$ for $x>0.$

$y=\dfrac{200-4x^3}{\pi x^2}=x\cdot\dfrac{200-4x^3}{\pi x^3}$

$=\sqrt[3]{\dfrac{200}{4+\pi}}\cdot\dfrac{200-4(\dfrac{200}{4+\pi})}{\pi (\dfrac{200}{4+\pi})}$

$=\sqrt[3]{\dfrac{200}{4+\pi}}\cdot\dfrac{4+\pi-4}{\pi\\}=\sqrt[3]{\dfrac{200}{4+\pi}}$

The length of the cube is $\sqrt[3]{\dfrac{200}{4+\pi}}\ cm\approx3.037\ cm.$

The diameter of the base of the cylinder is $\sqrt[3]{\dfrac{200}{4+\pi}}\ cm\approx3.037\ cm.$

The height of the cylinder is $\sqrt[3]{\dfrac{200}{4+\pi}}\ cm\approx3.037\ cm.$