To solve the problem of finding the dimensions of a box with a volume of 20 cubic feet that minimizes the cost, we can use optimization techniques. The goal is to minimize the cost function while ensuring the box maintains a volume of 20 cubic feet.

1. Define the Variables:

Let:

• $x$ = the length of the box (in feet),
• $y$ = the width of the box (in feet),
• $h$ = the height of the box (in feet).

2. Volume Constraint:

The volume of the box is given by: $V = x \cdot y \cdot h = 20 \, \text{cubic feet}$ This is our volume constraint.

3. Cost Function:

The cost of the box includes the costs of the sides, lid, and bottom:

• The sides (4 sides) and the lid are made of material that costs $0.20 per square foot.
• The bottom is made of material that costs $0.30 per square foot.

The surface area components are:

• The area of the bottom = $x \cdot y$,
• The area of the lid = $x \cdot y$,
• The area of the four sides = $2(x \cdot h + y \cdot h)$.

Thus, the total cost $C$ is: $C = 0.30(xy) + 0.20(xy) + 0.20[2(xh + yh)]$ Simplifying the cost function: $C = 0.50(xy) + 0.40(xh + yh)$

4. Solve for $h$:

Using the volume constraint $x \cdot y \cdot h = 20$, we can express $h$ as: $h = \frac{20}{xy}$

5. Substitute $h$ into the Cost Function:

Substituting $h = \frac{20}{xy}$ into the cost function: $C = 0.50(xy) + 0.40\left[x \cdot \frac{20}{xy} + y \cdot \frac{20}{xy}\right]$ Simplifying the terms: $C = 0.50(xy) + 0.40\left[\frac{20}{y} + \frac{20}{x}\right]$ $C = 0.50(xy) + 8\left(\frac{1}{x} + \frac{1}{y}\right)$

6. Minimize the Cost Function:

To minimize the cost, we take the partial derivatives of $C$ with respect to $x$ and $y$, set them equal to zero, and solve for $x$ and $y$.

Partial derivative with respect to $x$:

$\frac{\partial C}{\partial x} = 0.50y - 8\frac{1}{x^2} = 0$ $0.50y = \frac{8}{x^2}$ $y = \frac{16}{x^2}$

Partial derivative with respect to $y$:

$\frac{\partial C}{\partial y} = 0.50x - 8\frac{1}{y^2} = 0$ $0.50x = \frac{8}{y^2}$ $x = \frac{16}{y^2}$

7. Solve the System of Equations:

Substitute $y = \frac{16}{x^2}$ into the volume constraint $x \cdot y \cdot h = 20$.

Solving these equations will give us the values of $x$, $y$, and $h$ that minimize the cost.

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