question 3 (10 points). a box with a square base and an open top must have a volume of 32,000 cm³. find the…

question 3 (10 points). a box with a square base and an open top must have a volume of 32,000 cm³. find the dimensions of the box that minimize the amount of material used.
Answer
Explanation:
Step1: Define variables
Let the side - length of the square base be $x$ (in cm) and the height of the box be $h$ (in cm). The volume $V$ of the box is $V=x^{2}h$, and we know that $V = 32000$, so $h=\frac{32000}{x^{2}}$. The surface area $S$ of the open - top box is $S=x^{2}+4xh$.
Step2: Substitute $h$ into the surface - area formula
Substitute $h=\frac{32000}{x^{2}}$ into $S$: $S(x)=x^{2}+4x\cdot\frac{32000}{x^{2}}=x^{2}+\frac{128000}{x},x>0$.
Step3: Find the derivative of $S(x)$
Using the power rule, if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. $S^\prime(x)=2x-\frac{128000}{x^{2}}$.
Step4: Set the derivative equal to zero and solve for $x$
Set $S^\prime(x) = 0$: $2x-\frac{128000}{x^{2}}=0$. Multiply through by $x^{2}$ to get $2x^{3}-128000 = 0$. $2x^{3}=128000$, then $x^{3}=64000$, and $x = 40$.
Step5: Find the second - derivative of $S(x)$
$S^{\prime\prime}(x)=2+\frac{256000}{x^{3}}$. When $x = 40$, $S^{\prime\prime}(40)=2+\frac{256000}{40^{3}}=2 + 4=6>0$. So $S(x)$ has a minimum at $x = 40$.
Step6: Find the value of $h$
Substitute $x = 40$ into the formula for $h$: $h=\frac{32000}{40^{2}}=\frac{32000}{1600}=20$.
Answer:
The side - length of the square base is $40$ cm and the height is $20$ cm.