A box is to be made where the material for the sides and the lid cost​ $0.20 per square foot and the cost for the bottom is ​$0.30 per square foot. Find the dimensions of a box with volume 20 cubic feet that has minimum cost.

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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A box is to be made where the material for the sides and the lid cost $0.20 per square foot and the cost for the bottom is $0.30 per square foot. Find the dimensions of a box with volume 20 cubic feet that has minimum cost.

Learn how to find the dimensions of a box with a volume of 20 cubic feet that minimizes cost using optimization techniques. This solution includes concepts like partial derivatives, volume constraints, and cost functions, ideal for students studying calculus.

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