question: you have been asked to design a rectangular box with a square base and an open - top. the volume…
question: you have been asked to design a rectangular box with a square base and an open - top. the volume of the box must be 24 cm³. determine the minimum surface area necessary to construct a box of this volume. enter an exact answer. provide your answer below: □ cm²
Answer
Explanation:
Step1: Let the side - length of the square base be $x$ cm and the height of the box be $h$ cm.
The volume $V=x^{2}h$. Given $V = 24$, so $h=\frac{24}{x^{2}}$.
Step2: Find the surface - area formula.
The surface area $A$ of the open - top box with a square base is $A=x^{2}+4xh$. Substitute $h = \frac{24}{x^{2}}$ into the surface - area formula: $A=x^{2}+4x\cdot\frac{24}{x^{2}}=x^{2}+\frac{96}{x},x\gt0$.
Step3: Differentiate the surface - area function.
Differentiate $A(x)$ with respect to $x$. Using the power rule, $A^\prime(x)=2x-\frac{96}{x^{2}}$.
Step4: Find the critical points.
Set $A^\prime(x) = 0$. Then $2x-\frac{96}{x^{2}}=0$. Multiply through by $x^{2}$ to get $2x^{3}-96 = 0$. So $x^{3}=48$, and $x=\sqrt[3]{48}=2\sqrt[3]{6}$.
Step5: Find the second - derivative.
Differentiate $A^\prime(x)$ to get the second - derivative $A^{\prime\prime}(x)=2+\frac{192}{x^{3}}$. When $x = 2\sqrt[3]{6}$, $A^{\prime\prime}(x)\gt0$, which means the function $A(x)$ has a minimum at $x = 2\sqrt[3]{6}$.
Step6: Calculate the minimum surface area.
Substitute $x = 2\sqrt[3]{6}$ into the surface - area formula $A(x)$. [ \begin{align*} A(2\sqrt[3]{6})&=(2\sqrt[3]{6})^{2}+\frac{96}{2\sqrt[3]{6}}\ &=4\sqrt[3]{36}+\frac{48}{\sqrt[3]{6}}\ &=4\sqrt[3]{36}+8\sqrt[3]{144}\ &=4\sqrt[3]{36}+8\sqrt[3]{6^{2}\times4}\ &=4\sqrt[3]{36}+16\sqrt[3]{6} \end{align*} ]