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
[tex]x^2\cdot y = 48668[/tex]. From here, we know that [tex]y=\frac{48668}{x^2}[/tex]
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 [tex]x^2[/tex]. Recall that each side is a rectangle of base x and height y. So, the are of one side is [tex]x\cdot y[/tex]. Then, the area of the box is given by the function
[tex]A(x,y) = x^2+4xy[/tex]. Since we can relate y to x through the volume, this can be a function of one variable. Namely
[tex]A(x) = x^2+4x\cdot \frac{48668}{x^2}= x^2+4\cdot \frac{48668}{x} = x^2+48668x^{-1}[/tex]
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 [tex]x^n[/tex] is [tex]nx^{n-1}[/tex]. Then, applying the properties of derivatives, we have
[tex] A'(x) = 2x-4\cdot \frac{48668}{x^2} = \frac{2x^3-4\cdot 48668}{x^2}[/tex]
So, we want to solve the following equation
[tex]2x^3-4\cdot 48668 =0[/tex]
which implies [tex]x^3 = 97336[/tex]. Using a calculator we get the value [tex] x= 46[/tex]
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 [tex]A''(x) = 2+8\cdot\cdot 48668}{x^3}[/tex], note that A''(46) = 6>0, so x=46 is the minimum value of the area.
