Answer:
[tex]V_m=3.74\ ft^2[/tex]
Step-by-step explanation:
Maximization Using Derivatives
We'll use the first derivative to find the extreme values of a function and then find the value of the independent variables to make the function maximum.
The first step is to produce a model that contains only one variable, take its first derivative and equal it to 0.
Assume a box with a square bottom and rectangular sides. Let's set the sides of the bottom square as x and the height of the box as y. The volume of the box is
[tex]V=x^2y[/tex]
And the total side area is the sum of the bottom square plus the 4 side areas, each one being a rectangle of dimensions x and y. The area is
[tex]A=x^2+4xy[/tex]
We know the total material used for the box is [tex]10 ft^2[/tex]
[tex]x^2+4xy=10[/tex]
Solving for y
[tex]\displaystyle y=\frac{10-x^2}{4x}[/tex]
The expression for the volume will now be a function only of x:
[tex]\displaystyle V=x^2\cdot \frac{10-x^2}{4x}[/tex]
[tex]\displaystyle V=\frac{10x-x^3}{4}[/tex]
Taking the derivative of V
[tex]\displaystyle V'=\frac{10-3x^2}{4}[/tex]
Equating to 0
[tex]\displaystyle \frac{10-3x^2}{4}=0[/tex]
Solving for x
[tex]\displaystyle x=\sqrt{\frac{10}{3}}=1.83[/tex]
[tex]x=1.83\ ft[/tex]
Computing the second derivative:
[tex]\displaystyle V''=\frac{-3x}{2}[/tex]
The second derivative is negative for any positive value of x, thus the extreme value of x is a maximum, and the largest volume of the box is
[tex]\displaystyle V_m=\frac{10\cdot 1.83-1.83^3}{4}[/tex]
[tex]V_m=3.74\ ft^2[/tex]
