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find the dimensions of a closed rectangular box with a square base and …

Question

find the dimensions of a closed rectangular box with a square base and volume 27 in³ that can be constructed with the least amount of material. what are the dimensions of the box? the length of one side of the base is in. the height of the box is in.

Explanation:

Step1: Define variables

Let the side - length of the square base be $x$ (in inches) and the height of the box be $h$ (in inches). The volume $V$ of the box is given by $V=x^{2}h$, and $V = 27$, so $h=\frac{27}{x^{2}}$. The surface area $S$ of the closed rectangular box with a square base is $S = 2x^{2}+4xh$.

Step2: Substitute $h$ into the surface - area formula

Substitute $h=\frac{27}{x^{2}}$ into $S$: $S = 2x^{2}+4x\cdot\frac{27}{x^{2}}=2x^{2}+\frac{108}{x},x>0$.

Step3: Find the derivative of $S$ with respect to $x$

Using the power rule, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. The derivative $S^\prime(x)=4x-\frac{108}{x^{2}}$.

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

Set $S^\prime(x)=0$:
\[

$$\begin{align*} 4x-\frac{108}{x^{2}}&=0\\ 4x&=\frac{108}{x^{2}}\\ 4x^{3}&=108\\ x^{3}& = 27\\ x&=3 \end{align*}$$

\]

Step5: Find the second - derivative of $S$ to confirm it's a minimum

The second - derivative $S^{\prime\prime}(x)=4+\frac{216}{x^{3}}$. When $x = 3$, $S^{\prime\prime}(3)=4+\frac{216}{27}=4 + 8=12>0$. So $x = 3$ gives a minimum of the surface - area function.

Step6: Find the value of $h$

Substitute $x = 3$ into the formula for $h$: $h=\frac{27}{3^{2}}=\frac{27}{9}=3$.

Answer:

The length of one side of the base is 3 in.
The height of the box is 3 in.