Question

A box with a square base and open top must have a volume of 48668 cm3. We wish to find the dimensions of the box that minimize the amount of material used.First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base.[Hint: use the volume formula to express the height of the box in terms of x.]Simplify your formula as much as possible.

Answer 1

Answer:

x=46 gives us the minimum value

Step-by-step explanation:

Let us have the height of the box to be y. Since the box has a square base, let us call the side of the base as x. So, the volume of the box is given by

$x^2\cdot y = 48668$. From here, we know that $y=\frac{48668}{x^2}$

Now, we will find the area of the box. Since the box has an open top, then we only have the 4 sides of the box and the base. The base is a square of side x, so its area is $x^2$. Recall that each side is a rectangle of base x and height y. So, the are of one side is $x\cdot y$. Then, the area of the box is given by the function

$A(x,y) = x^2+4xy$. Since we can relate y to x through the volume, this can be a function of one variable. Namely

$A(x) = x^2+4x\cdot \frac{48668}{x^2}= x^2+4\cdot \frac{48668}{x} = x^2+48668x^{-1}$

If we want to find the mininum value, we should derive the function A and find the value of x for which A'(x) is zero. REcall that the derivative of a function of the form $x^n$ is $nx^{n-1}$. Then, applying the properties of derivatives, we have

$A'(x) = 2x-4\cdot \frac{48668}{x^2} = \frac{2x^3-4\cdot 48668}{x^2}$

So, we want to solve the following equation

$2x^3-4\cdot 48668 =0$

which implies $x^3 = 97336$. Using a calculator we get the value $x= 46$

REcall that to check if the point is a minimum, we use the second derivative criteri. It states that the point x is a mininimum if f'(x) =0 and f''(x)>0

Note that $A''(x) = 2+8\cdot\cdot 48668}{x^3}$, note that A''(46) = 6>0, so x=46 is the minimum value of the area.