An aquarium is being built in the shape of a rectangular box with a square bottom and no top.

Let represent the length of one side of the bottom of the aquarium and let represent the height of the aquarium.

Suppose we wish to minimize the amount of material needed to create the aquarium. If the finished aquarium must hold 32 ft of water, what is the equation we should optimize?

Question

An aquarium is being built in the shape of a rectangular box with a square bottom and no top.

Let  represent the length of one side of the bottom of the aquarium and let   represent the height of the aquarium.

Suppose we wish to minimize the amount of material needed to create the aquarium. If the finished aquarium must hold 32 ft  of water, what is the equation we should optimize?

BlackTom AI · Accepted Answer

First, introduce variables consistent with the problem: let the square base have side length x feet and the height be y feet.
Next, write the volume constraint for the aquarium. Since the base is square, the volume V is V = base area × height = x^2 × y, and we’re given V = 32 ft^3. This yields a relation between x and y: y = 32 / x^2.
Now express the surface area to be minimized. The aquarium has a square bottom and four side faces (no top). So the surface area S consists of the bottom area plus the lateral area: S = (base area) + (lateral area) = x^2 + (4 × x × y).
Substitute the height expression from the volume constraint into the surface area: S(x) = x^2 + 4x(32 / x^2) = x^2 + 128/x.
To find the minimization, analyze S as a function of x > 0. Compute the derivative: S'(x) = 2x − 128/x^2. Set the derivative equal to zero to locate critical points: 2x − 128/x^2 = 0.
Multiply both sides by x^2 to clear the denominator: 2x^3 − 128 = 0, which simplifies to x^3 = 64. From here you would determine x by taking the cube root (ensuring x > 0).
Once x is found, use the volume constraint to determine y: y = 32 / x^2. These steps yield the dimensions that minimize the material used, and the corresponding minimal surface area can then be computed by substituting x and y back into S = x^2 + 4xy.
In summary, the equation to optimize is S(x) = x^2 + 128/x with x > 0, derived by substituting y = 32/x^2 into the surface-area expression S = x^2 + 4xy, and the minimization proceeds by solving S'(x) = 0.

MATH_1225_17255_202501 4.7 Optimization

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