Question

1. a rectangular open - topped aquarium is to have a square base and volume 8 m³. the material for the base costs $1 per m², and the material for the sides costs $4 per m². what dimensions minimize the cost of the aquarium? (make sure to justify why your answer corresponds to an absolute minimum.)

Explanation:

Step1: Define variables

Let the side length of the square base be \(x\) (in meters) and the height be \(h\) (in meters). The volume \(V=x^{2}h\), and since \(V = 8\), we have \(h=\frac{8}{x^{2}}\).

The cost function \(C\): The area of the base is \(A_{base}=x^{2}\), and the cost for the base is \(1\times x^{2}\). The area of the four - sides is \(A_{sides}=4xh\), and the cost for the sides is \(4\times4xh\). So \(C(x)=x^{2}+16xh\).

Substitute \(h = \frac{8}{x^{2}}\) into the cost function: \(C(x)=x^{2}+16x\times\frac{8}{x^{2}}=x^{2}+\frac{128}{x},x>0\).

Step2: Find the derivative of the cost function

Differentiate \(C(x)\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(C^\prime(x)=2x-\frac{128}{x^{2}}\).

Set \(C^\prime(x) = 0\) for critical points:

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Step3: Find the second - derivative

Differentiate \(C^\prime(x)\) to get the second - derivative. \(C^{\prime\prime}(x)=2+\frac{256}{x^{3}}\).

When \(x = 4\), \(C^{\prime\prime}(4)=2+\frac{256}{64}=2 + 4=6>0\).

Since \(C^{\prime\prime}(x)>0\) for \(x = 4\), the function \(C(x)\) has a local minimum at \(x = 4\).

As \(x>0\) (the domain of the function \(C(x)\) based on the physical meaning of length), and the function \(C(x)\) has only one critical point in its domain, this local minimum is an absolute minimum.

When \(x = 4\), \(h=\frac{8}{x^{2}}=\frac{8}{16}=\frac{1}{2}\)

Answer:

The base has a side length of \(4\) meters and the height is \(\frac{1}{2}\) meter.