Question Statement

An open rectangular box is to be constructed with a square base and a volume of $32,000{\mathrm{cm}}^3$. Find the dimensions of box that require the least amount of material.

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Background and Explanation

This problem requires optimization using differential calculus. We need to minimize the surface area (material used) of an open-top box subject to a fixed volume constraint. The approach involves expressing the surface area as a function of a single variable, then using derivatives to find the minimum.

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Solution

Let the side length of the square base be $x$ and the height be $y$.

Step 1: Establish the constraint equation from the volume

Given that the volume of the box is $32,000{\mathrm{cm}}^3$:

$\begin{array}{rl}\text{Volume}& =x\cdot x\cdot y=32000\\ x^{2}y& =32000\end{array}$

Solving for $y$ in terms of $x$:

$y=\frac{32000}{x^2}$

Step 2: Formulate the surface area function

Since the box is open at the top, the material is used for the square base and the four rectangular walls:

• Area of base: $x^2$
• Area of one wall: $xy$
• Area of four walls: $4xy$

Therefore, the total surface area $A$ is:

$A=x^2+4xy$

Substituting equation (1) to express $A$ as a function of $x$ only:

$\begin{array}{rl}A& =x^2+4x\left(\frac{32000}{x^2}\right)\\ A& =x^2+\frac{128000}{x}\end{array}$

Step 3: Find the critical points

Differentiate $A$ with respect to $x$:

$\begin{array}{rl}\frac{dA}{dx}& =2x+128000\cdot \frac{d}{dx}\left(\frac{1}{x}\right)\\ & =2x+128000\left(-\frac{1}{x^2}\right)\\ \frac{dA}{dx}& =2x-\frac{128000}{x^2}\end{array}$

Set $\frac{dA}{dx}=0$ to find critical values:

$\begin{array}{rl}2x-\frac{128000}{x^2}& =0\\ 2x& =\frac{128000}{x^2}\\ 2x^3& =128000\\ x^3& =64000\\ x& =\sqrt[3]{64000}\\ x& =40\mathrm{cm}\end{array}$

Step 4: Verify this is a minimum using the second derivative test

Differentiate $\frac{dA}{dx}$ again with respect to $x$:

$\begin{array}{rl}\frac{d^{2}A}{d{x}^2}& =2-128000\cdot \frac{d}{dx}\left(x^{-2}\right)\\ & =2-128000\left(-2x^{-3}\right)\\ & =2+\frac{256000}{x^3}\\ & =2\left(1+\frac{128000}{x^3}\right)\end{array}$

Evaluate at $x=40$:

$\begin{array}{rl}\frac{d^{2}A}{d{x}^2}& =2\left(1+\frac{128000}{(40{)}^3}\right)\\ & =2\left(1+\frac{128000}{64000}\right)\\ & =2(1+2)\\ & =6>0\end{array}$

Since $\frac{d^{2}A}{d{x}^2}>0$ at $x=40$, the surface area is indeed minimized at this value.

Step 5: Calculate the corresponding height

Substitute $x=40$ into equation (1):

$\begin{array}{rl}y& =\frac{32000}{(40{)}^2}\\ & =\frac{32000}{1600}\\ & =20\mathrm{cm}\end{array}$

Final Answer:

• Length of base: $x=40\mathrm{cm}$
• Height: $y=20\mathrm{cm}$

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Key Formulas or Methods Used

• Volume of rectangular box: $V=\text{length}\times \text{width}\times \text{height}=x^{2}y$ (since base is square)
• Surface area of open box: $A=\text{base area}+\text{4 walls}=x^2+4xy$
• First derivative test: Set $\frac{dA}{dx}=0$ to find critical points
• Second derivative test for minimum: If $\frac{d^{2}A}{d{x}^2}>0$ at a critical point, the function has a local minimum there
• Power rule for differentiation: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

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Summary of Steps

1. Define variables: Let $x$ be the side of the square base and $y$ be the height
2. Apply volume constraint: Use $x^{2}y=32000$ to express $y=\frac{32000}{x^2}$
3. Create objective function: Write surface area $A=x^2+4xy$ and substitute to get $A(x)=x^2+\frac{128000}{x}$
4. Differentiate: Find $\frac{dA}{dx}=2x-\frac{128000}{x^2}$
5. Find critical points: Solve $\frac{dA}{dx}=0$ to get $x=40$ cm
6. Verify minimum: Check $\frac{d^{2}A}{d{x}^2}=6>0$ at $x=40$ to confirm it's a minimum
7. Find height: Calculate $y=\frac{32000}{40^2}=20$ cm
8. State dimensions: The optimal dimensions are $40\mathrm{cm}\times 40\mathrm{cm}\times 20\mathrm{cm}$