The volume of the box will be given by:
[tex]V=x(20-2x)(40-2x)[/tex]
And the according graph is:
To find the value of x that provide the maximum value of the box, we could find the deritative of the volume function:
[tex]V=x(20-2x)(40-2x)[/tex]
First, let's rewrite the function as:
[tex]V(x)=4x^3-120x^2+800x[/tex]
The deritative of this function is:
[tex]\frac{dV}{dx}=12x^2-240x+800[/tex]
Now, we're going to equal this equation to zero and then solve for x:
[tex]\begin{gathered} \frac{dV}{dx}=0 \\ \\ 12x^2-240x+800=0 \\ Solving\colon \\ x=\begin{cases}\frac{30-10\sqrt[]{3}}{3} \\ \frac{30+10\sqrt[]{3}}{3}\end{cases} \end{gathered}[/tex]
We obtained two solutions which are the critical points of V. The first solution is the point that shows a maximum in the graph of V(x), so, the value of x that provide the maximum value of the graph is:
[tex]x=\frac{30-10\sqrt[]{3}}{3}[/tex]
