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Answer:
B. 'x' should be 1.53 inches
Step-by-step explanation:
Given equation for the volume of the box:
[tex]V=(8-2x)(11-2x)(x)[/tex]
To find the value of x that will maximize the volume, differentiate the volume, set the first derivative to zero and solve for x.
Expand the equation:
[tex]V=x(8-2x)(11-2x)[/tex]
[tex]\implies V=x(88-16x-22x+4x^2)[/tex]
[tex]\implies V=4x^3-38x^2+88x[/tex]
Differentiate:
[tex]\implies \dfrac{dV}{dx}=12x^2-76x+88[/tex]
Set to zero and solve for x:
[tex]\implies \dfrac{dV}{dx}=0[/tex]
[tex]\implies 12x^2-76x+88=0[/tex]
Divide by 4:
[tex]\implies 3x^2-19x+22=0[/tex]
Use the quadratic formula to solve the quadratic equation.
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]
[tex]\implies x=\dfrac{-(-19)\pm \sqrt{(-19)^2-4(3)(22)}}{2(3)}[/tex]
[tex]\implies x=\dfrac{19\pm \sqrt{97}}{6}[/tex]
To find which value of x maximizes the volume, find the second derivative:
[tex]\implies \dfrac{d^2V}{dx^2}=24x-76[/tex]
Then input the values of x into the second derivative:
[tex]x=\dfrac{19+\sqrt{97}}{6} \implies \dfrac{d^2V}{dx^2}=4\sqrt{97} > 0\implies \textsf{minimum}[/tex]
[tex]x=\dfrac{19-\sqrt{97}}{6} \implies \dfrac{d^2V}{dx^2}=-4\sqrt{97} < 0\implies \textsf{maximum}[/tex]
Therefore, the value of x that will maximize the volume is:
[tex]x=\dfrac{19-\sqrt{97}}{6}=1.53\:\sf inches \:(nearest\:hundredth)[/tex]
Alternatively, you can input the given options of x into the formula and compare results, but this is the correct way to find/prove it.