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.
sorry, thats incorrect. try again?
4√3{36}+16√3{6} cm²

Explanation:

Step1: Define variables

Let the side - length of the square base be $x$ cm and the height of the box be $h$ cm. The volume $V=x^{2}h$, and we know $V = 24$ cm³, so $h=\frac{24}{x^{2}}$.

Step2: Find the surface - area formula

The surface area $A$ of an open - top box with a square base is $A=x^{2}+4xh$. Substitute $h=\frac{24}{x^{2}}$ into the surface - area formula: $A=x^{2}+4x\cdot\frac{24}{x^{2}}=x^{2}+\frac{96}{x}$, where $x>0$.

Step3: Find the derivative of the surface - area function

Differentiate $A(x)$ with respect to $x$. Using the power rule, if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. So $A^\prime(x)=2x-\frac{96}{x^{2}}$.

Step4: Set the derivative equal to zero and solve for $x$

Set $A^\prime(x)=0$, then $2x-\frac{96}{x^{2}} = 0$. Multiply through by $x^{2}$ to get $2x^{3}-96 = 0$. Rearrange to $x^{3}=48$, so $x=\sqrt[3]{48}=2\sqrt[3]{6}$ cm.

Step5: Find the second - derivative of the surface - area function

Differentiate $A^\prime(x)$ to get the second - derivative $A^{\prime\prime}(x)=2+\frac{192}{x^{3}}$. When $x = 2\sqrt[3]{6}$, $A^{\prime\prime}(x)>0$, which means the surface area has a minimum at this $x$ value.

Step6: Find the minimum surface area

Substitute $x = 2\sqrt[3]{6}$ into the surface - area formula $A(x)$.
\[

$$\begin{align*} A&=(2\sqrt[3]{6})^{2}+\frac{96}{2\sqrt[3]{6}}\\ &=4\sqrt[3]{36}+\frac{48}{\sqrt[3]{6}}\\ &=4\sqrt[3]{36}+8\sqrt[3]{144}\\ &=4\sqrt[3]{36}+16\sqrt[3]{6} \end{align*}$$

\]

Answer:

$4\sqrt[3]{36}+16\sqrt[3]{6}\text{ cm}^2$