Question

a pizza box with a lid is made from a 19-inch-by-40-inch piece of cardboard. the volume of the pizza box is represented by the equation below, where x is the height of the box. what height will give the maximum volume for the pizza box?
$v(x) = \frac{1}{2}x(19 - 2x)(40 - 5x)$
○ 976.16
○ 2.88
○ 3.65
○ 1.79

Explanation:

Step1: Expand volume function

First, expand $V(x) = \frac{1}{2}x(19-2x)(40-5x)$:
First multiply $(19-2x)(40-5x) = 19*40 -19*5x -2x*40 + 10x^2 = 760 -95x -80x +10x^2 = 10x^2 -175x +760$
Then multiply by $\frac{1}{2}x$:
$V(x) = \frac{1}{2}x(10x^2 -175x +760) = 5x^3 - \frac{175}{2}x^2 + 380x$

Step2: Find first derivative

Take derivative of $V(x)$ to find critical points:
$V'(x) = 15x^2 - 175x + 380$

Step3: Solve quadratic equation

Set $V'(x)=0$, solve $15x^2 -175x +380=0$. Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=15$, $b=-175$, $c=380$:
First calculate discriminant:
$\Delta = (-175)^2 -4*15*380 = 30625 - 22800 = 7825$
$\sqrt{\Delta} = \sqrt{7825} \approx 88.46$
Then:
$x_1 = \frac{175 + 88.46}{30} \approx \frac{263.46}{30} \approx 8.78$
$x_2 = \frac{175 - 88.46}{30} \approx \frac{86.54}{30} \approx 2.88$

Step4: Validate feasible height

Check if values make sense for the box:
For $x=8.78$, $19-2x=19-17.56=1.44$, $40-5x=40-43.9=-3.9$ (negative, invalid, since length can't be negative)
For $x=2.88$, $19-2x=19-5.76=13.24$, $40-5x=40-14.4=25.6$ (both positive, valid)

Step5: Confirm maximum

Second derivative test: $V''(x)=30x-175$. For $x=2.88$, $V''(2.88)=30*2.88-175=86.4-175=-88.6<0$, so this is a maximum.

Answer:

2.88