Question

All boxes with a square​ base, an open​ top, and a volume of 60 ft cubed have a surface area given by ​S(x)equalsx squared plus StartFraction 240 Over x EndFraction ​, where x is the length of the sides of the base. Find the absolute minimum of the surface area function on the interval ​(0,infinity​). What are the dimensions of the box with minimum surface​ area?

Answer 1

The absolute minimum of the surface area function on the interval
`(0,\infty)` is
`S(2\sqrt[3]{15})=12\cdot \:15^{(2)/(3)} \:ft^2`

The dimensions of the box with minimum surface​ area are: the base edge
`x=2\sqrt[3]{15}\:ft` and the height
`h=\sqrt[3]{15} \:ft`

Explanation:

We are given the surface area of a box
`S(x)=x^2+(240)/(x)` where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval
`(0,\infty)` and the dimensions of the box with minimum surface​ area.

1. To find the absolute minimum you must find the derivative of the surface area (
`S'(x)`) and find the critical points of the derivative (
`S'(x)=0`).

`(d)/(dx) S(x)=(d)/(dx)(x^2+(240)/(x))\\\\(d)/(dx) S(x)=(d)/(dx)\left(x^2\right)+(d)/(dx)\left((240)/(x)\right)\\\\S'(x)=2x-(240)/(x^2)`

Next,

`2x-(240)/(x^2)=0\\2xx^2-(240)/(x^2)x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120`

There is a undefined solution
`x=0` and a real solution
`x=2\sqrt[3]{15}`. These point divide the number line into two intervals
`(0,2\sqrt[3]{15})` and
`(2\sqrt[3]{15}, \infty)`

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

`\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&(16)/(3)&increasing \end{array}`

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before
`x=2\sqrt[3]{15}`, increases after it, and is defined at
`x=2\sqrt[3]{15}`. So f(x) has a relative minimum point at
`x=2\sqrt[3]{15}`.

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for
`x=2\sqrt[3]{15}`.

`(d)/(dx) S'(x)=(d)/(dx)(2x-(240)/(x^2))\\\\S''(x) =(d)/(dx)\left(2x\right)-(d)/(dx)\left((240)/(x^2)\right)\\\\S''(x) =2+(480)/(x^3)`

and for
`x=2\sqrt[3]{15}` we get:

`2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0`

Therefore S(x) has a minimum at
`x=2\sqrt[3]{15}` which is:

`S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{(2)/(3)}+2^3\cdot \:15^{(2)/(3)}\\\\4\cdot \:15^{(2)/(3)}+8\cdot \:15^{(2)/(3)}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{(2)/(3)} \:ft^2`

2. To find the third dimension of the box with minimum surface​ area:

We know that the volume is 60
`ft^3` and the volume of a box with a square base is
`V=x^2h`, we solve for h

`h=(V)/(x^2)`

Substituting V = 60
`ft^3` and
`x=2\sqrt[3]{15}`

`h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{(2)/(3)}}\\\\h=\sqrt[3]{15} \:ft`

The dimension are the base edge
`x=2\sqrt[3]{15}\:ft` and the height
`h=\sqrt[3]{15} \:ft`