You are to construct an open rectangular box with a square base and a volume of 96 ft³. If material for the bottom costs $3 per sq. ft and material for the sides costs $1 per sq. f... You are to construct an open rectangular box with a square base and a volume of 96 ft³. If material for the bottom costs $3 per sq. ft and material for the sides costs $1 per sq. ft, what dimensions will result in the least expensive box? What is the minimum cost?
Understand the Problem
The question is asking for the dimensions of an open rectangular box with a square base that has a specific volume, in order to minimize the cost of materials for its construction. We will approach this by using optimization techniques, specifically through setting up a function for cost based on the dimensions and finding its minimum value.
Answer
The least expensive box dimensions are $4 \, ft$ (base side) and $6 \, ft$ (height), with a cost of $144$.
Answer for screen readers
The dimensions that result in the least expensive box are a square base with a side length of $4 , ft$ and a height of $6 , ft$. The minimum cost is $144$ dollars.
Steps to Solve
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Define Variables Let the length of the side of the square base be $x$ (in feet), and the height of the box be $h$ (in feet).
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Set up the Volume Constraint Given that the volume $V$ of the box is $96 , ft^3$, we have the equation: $$ V = x^2 h = 96 $$
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Express Height in Terms of Base Length From the volume equation, solve for $h$: $$ h = \frac{96}{x^2} $$
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Set up the Cost Function The cost $C$ consists of the cost of the bottom and the sides. The area of the bottom is $x^2$, and the area of the sides is $4xh$: $$ C = 3x^2 + 4(1)(xh) = 3x^2 + 4x\left(\frac{96}{x^2}\right) $$
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Simplify the Cost Function Substituting for $h$, the cost function becomes: $$ C = 3x^2 + \frac{384}{x} $$
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Differentiate the Cost Function To minimize the cost, differentiate $C$ with respect to $x$: $$ C' = 6x - \frac{384}{x^2} $$
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Find Critical Points Set the derivative equal to zero and solve for $x$: $$ 6x - \frac{384}{x^2} = 0 $$ $$ 6x^3 = 384 $$ $$ x^3 = 64 $$ $$ x = 4 $$
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Determine the Height Now substitute $x = 4$ back into the equation for $h$: $$ h = \frac{96}{4^2} = \frac{96}{16} = 6 $$
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Conclusion with Dimensions The dimensions of the box are:
- Side length of base: $x = 4 , ft$
- Height: $h = 6 , ft$
- Calculate Minimum Cost Substituting $x = 4$ into the cost function: $$ C = 3(4^2) + 4(4)(6) $$ $$ C = 3(16) + 96 = 48 + 96 = 144 $$
The dimensions that result in the least expensive box are a square base with a side length of $4 , ft$ and a height of $6 , ft$. The minimum cost is $144$ dollars.
More Information
The box's design is an application of optimization techniques commonly used in calculus. Given the constraints of the materials and costs, this problem exemplifies how to use derivatives to find cost-effective solutions in practical scenarios.
Tips
- Forgetting to substitute back for height after finding the base length.
- Not properly taking the derivative or setting it to zero to find the critical points.
- Confusing the cost per square foot for the sides and bottom, leading to incorrect formulations of the cost function.
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