AP CALCULUS - Optimization Problems Practice | PDF | Area | Rectangle - Free Printable

Problem 1: Maximizing Profit for Smartphone Manufacturing

#### Given:
- Price per smartphone: \( P(x) = 110 - 0.05x \)
- Cost per smartphone: $50
- Fixed daily cost: $6000
- Profit function: \( \text{Profit} = \text{Revenue} - \text{Cost} \)

#### Step-by-Step Solution:
1. Define Revenue Function:
The revenue \( R(x) \) is the price per smartphone multiplied by the number of smartphones sold:
\[
R(x) = x \cdot (110 - 0.05x) = 110x - 0.05x^2
\]

2. Define Cost Function:
The total cost \( C(x) \) includes the variable cost per smartphone and the fixed daily cost:
\[
C(x) = 50x + 6000
\]

3. Define Profit Function:
The profit \( P(x) \) is the difference between revenue and cost:
\[
P(x) = R(x) - C(x) = (110x - 0.05x^2) - (50x + 6000)
\]
Simplify:
\[
P(x) = 110x - 0.05x^2 - 50x - 6000 = -0.05x^2 + 60x - 6000
\]

4. Maximize the Profit Function:
The profit function \( P(x) = -0.05x^2 + 60x - 6000 \) is a quadratic function in the form \( ax^2 + bx + c \), where \( a = -0.05 \), \( b = 60 \), and \( c = -6000 \). Since \( a < 0 \), the parabola opens downwards, and the maximum value occurs at the vertex.

The \( x \)-coordinate of the vertex of a parabola \( ax^2 + bx + c \) is given by:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = -0.05 \) and \( b = 60 \):
\[
x = -\frac{60}{2(-0.05)} = \frac{60}{0.1} = 600
\]

5. Conclusion:
The company should manufacture and sell 600 smartphones per day to maximize profit.

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Problem 2: Maximizing Area of Two Adjacent Corrals

#### Given:
- Total fencing available: 200 ft
- Two identical rectangular corrals sharing one side

#### Step-by-Step Solution:
1. Define Variables:
Let:
- \( a \): length of each corral (parallel to the shared side)
- \( b \): width of each corral (perpendicular to the shared side)

2. Fencing Constraint:
The total fencing used is:
- 3 widths (\( 3b \)) because the two corrals share one width.
- 2 lengths (\( 2a \)) because there are two lengths on the outer sides.
Therefore:
\[
2a + 3b = 200
\]

3. Express One Variable in Terms of the Other:
Solve for \( a \) in terms of \( b \):
\[
2a = 200 - 3b \implies a = 100 - \frac{3b}{2}
\]

4. Area of the Corrals:
The total area \( A \) of the two corrals is:
\[
A = 2 \cdot (a \cdot b) = 2ab
\]
Substitute \( a = 100 - \frac{3b}{2} \):
\[
A = 2 \left( \left( 100 - \frac{3b}{2} \right) b \right) = 2 \left( 100b - \frac{3b^2}{2} \right) = 200b - 3b^2
\]

5. Maximize the Area Function:
The area function \( A(b) = 200b - 3b^2 \) is a quadratic function in the form \( -3b^2 + 200b \). Since the coefficient of \( b^2 \) is negative, the parabola opens downwards, and the maximum value occurs at the vertex.

The \( b \)-coordinate of the vertex is:
\[
b = -\frac{B}{2A} = -\frac{200}{2(-3)} = \frac{200}{6} = \frac{100}{3}
\]

6. Find Corresponding \( a \):
Substitute \( b = \frac{100}{3} \) back into \( a = 100 - \frac{3b}{2} \):
\[
a = 100 - \frac{3 \left( \frac{100}{3} \right)}{2} = 100 - \frac{100}{2} = 100 - 50 = 50
\]

7. Conclusion:
The dimensions of each corral that maximize the total area are:
\[
\boxed{a = 50 \text{ ft}, b = \frac{100}{3} \text{ ft}}
\]

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Problem 3: Minimizing the Product of Two Rational Numbers

#### Given:
- One number is positive and exactly 8 greater than the other negative number.
- Minimize the product of these two numbers.

#### Step-by-Step Solution:
1. Define Variables:
Let:
- \( x \): the positive number
- \( y \): the negative number
Given \( x = y + 8 \) and \( y < 0 \).

2. Express the Product:
The product \( P \) of the two numbers is:
\[
P = x \cdot y
\]
Substitute \( x = y + 8 \):
\[
P = (y + 8)y = y^2 + 8y
\]

3. Minimize the Product Function:
The product function \( P(y) = y^2 + 8y \) is a quadratic function in the form \( y^2 + 8y \). Since the coefficient of \( y^2 \) is positive, the parabola opens upwards, and the minimum value occurs at the vertex.

The \( y \)-coordinate of the vertex is:
\[
y = -\frac{B}{2A} = -\frac{8}{2(1)} = -4
\]

4. Find Corresponding \( x \):
Substitute \( y = -4 \) back into \( x = y + 8 \):
\[
x = -4 + 8 = 4
\]

5. Calculate the Minimum Product:
The product \( P \) is:
\[
P = x \cdot y = 4 \cdot (-4) = -16
\]

6. Conclusion:
The minimum product of the two numbers is:
\[
\boxed{-16}
\]

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Final Answers:

1. Smartphone Manufacturing: \(\boxed{600}\)
2. Corral Dimensions: \(\boxed{a = 50 \text{ ft}, b = \frac{100}{3} \text{ ft}}\)
3. Minimum Product: \(\boxed{-16}\)