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 of the aquarium be \(h\) (in meters).
The volume \(V=x^{2}h\), and since \(V = 8\), we have \(h=\frac{8}{x^{2}}\).

Step2: Cost function

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: Each side has an area of \(xh\), and there are 4 sides. So the area of the sides is \(4xh\), and the cost for the sides is \(4\times4xh = 16xh\).
Then \(C(x)=x^{2}+16x\cdot\frac{8}{x^{2}}=x^{2}+\frac{128}{x},x>0\).

Step3: First - derivative

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 \(C^\prime(x) = 0\):

$$
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Step4: Second - derivative

Differentiate \(C^\prime(x)\) with respect to \(x\):
\(C^{\prime\prime}(x)=2+\frac{256}{x^{3}}\).
When \(x = 4\), \(C^{\prime\prime}(4)=2+\frac{256}{64}=2 + 4=6>0\).

Step5: Find height

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.