Optimization Problems and Applications of Derivatives in Business Calculus

Optimization Problems

Introduction to Optimization

Optimization is a fundamental concept in calculus and business mathematics, involving the determination of maximum or minimum values of a function under given constraints. These problems are common in economics, engineering, and management, where resources must be allocated efficiently.

• Objective Function: The function to be maximized or minimized (e.g., profit, area, cost).

• Constraint Equation: An equation representing the limitations or requirements of the problem (e.g., budget, perimeter).

General Steps for Solving Optimization Problems

1. Understand the Problem: Read the problem carefully and, if possible, draw a diagram. Define variables relevant to the objective.

2. Formulate the Objective Function: Write an equation for the quantity to be optimized.

3. Write the Constraint Equation: Express the limitations of the problem as an equation.

4. Reduce to One Variable: Solve the constraint for one variable and substitute into the objective function.

5. Differentiate: Take the derivative of the objective function with respect to the remaining variable.

6. Find Critical Points: Set the derivative equal to zero and solve for the variable to find potential maxima or minima.

7. Verify and Solve: Use the second derivative test or analyze endpoints to confirm whether the critical point yields a maximum or minimum. Substitute back to find all required quantities.

Example 1: Maximizing a Product

Problem: Find the maximum of $Q = x y$ if $x + y = 2$.

• Constraint: $x + y = 2$

• Objective: $Q = x y$

• Solution: Solve for $y = 2 - x$, substitute into $Q$:

$Q = x (2 - x) = 2x - x^2$

• Take derivative: $\frac{dQ}{dx} = 2 - 2x$

• Set to zero: $2 - 2x = 0 \implies x = 1$

• Thus, $y = 1$ and maximum $Q = 1 \times 1 = 1$

Example 2: Minimizing a Sum of Squares

Problem: Find the minimum of $Q = x^2 + y^2$ if $x + y = 8$.

• Constraint: $x + y = 8$

• Objective: $Q = x^2 + y^2$

• Solution: $y = 8 - x$, so $Q = x^2 + (8 - x)^2 = x^2 + 64 - 16x + x^2 = 2x^2 - 16x + 64$

• Derivative: $\frac{dQ}{dx} = 4x - 16$

• Set to zero: $4x - 16 = 0 \implies x = 4$

• Thus, $y = 4$ and minimum $Q = 16 + 16 = 32$

Example 3: Maximizing Area with Cost Constraints

Problem: You have $320 to fence a rectangular garden. The side facing the street costs $6/ft, other sides $2/ft. Find the largest possible area.

• Let: $x$ = length parallel to street, $y$ = width

• Cost constraint: $6x + 2x + 4y = 320$ (since three sides at $2/ft$, one at $6/ft$)

• Area: $A = x y$

• Solution: Express $y$ in terms of $x$ using the constraint, substitute into $A$, and optimize as above.

Additional info: Full solution involves calculus as in previous examples.

Example 4: Maximizing Area with Fixed Perimeter

Problem: Build a rectangular pen with three parallel walls using 500 ft of fencing. What dimensions maximize area?

• Let: $x$ = length, $y$ = width

• Constraint: $2y + 3x = 500$

• Area: $A = x y$

• Solution: Solve for $y$, substitute, and optimize $A$.

Example 5: Maximizing Area with Multiple Cost Constraints

Problem: A farmer has $750 to build an E-shaped fence along a river for two identical pastures. Parallel side costs $6/ft, perpendicular sides $5/ft. Find dimensions for maximum area.

• Let: $x$ = length parallel to river, $y$ = width

• Constraint: $6x + 3 \times 5y = 750$

• Area: $A = 2 x y$

• Solution: Express $y$ in terms of $x$, substitute, and optimize $A$.

Applications of Derivatives to Business and Economics

Key Functions in Business Calculus

• Cost Function ($C(x)$): Total cost of producing $x$ units.

• Revenue Function ($R(x)$): Total revenue from selling $x$ units. If price per unit is $p(x)$, then $R(x) = x \cdot p(x)$.

• Profit Function ($P(x)$): $P(x) = R(x) - C(x)$

• Demand Equation ($p(x)$): Price as a function of quantity demanded.

• Marginal: Refers to the derivative, representing the rate of change (e.g., marginal cost is $C'(x)$).

Example 6: Minimizing Marginal Cost

Problem: Given $C(x) = x^3 - 6x^2 + 13x + 15$, find the minimum marginal cost.

• Marginal cost: $C'(x) = 3x^2 - 12x + 13$

• Set derivative of $C'(x)$ to zero: $C''(x) = 6x - 12$

• Set $6x - 12 = 0 \implies x = 2$

• Minimum marginal cost at $x = 2$ is $C'(2) = 3(4) - 12(2) + 13 = 12 - 24 + 13 = 1$

Example 7: Maximizing Revenue

Problem: $R(x) = 200 - \frac{1600}{x+8} - x$. Find $x$ for maximum revenue and the maximum revenue.

• Take derivative: $R'(x) = \frac{1600}{(x+8)^2} - 1$

• Set to zero: $\frac{1600}{(x+8)^2} = 1 \implies x + 8 = 40 \implies x = 32$

• Maximum revenue: $R(32) = 200 - \frac{1600}{40} - 32 = 200 - 40 - 32 = 128$

Example 8: Maximizing Revenue with a Quadratic Demand

Problem: $p(x) = \frac{1}{12}x^2 - 10x + 300$, $0 \leq x \leq 60$. Find $x$ and $p(x)$ that maximize revenue, and the maximum revenue.

• Revenue: $R(x) = x \cdot p(x) = x \left( \frac{1}{12}x^2 - 10x + 300 \right )$

• Take derivative, set to zero, solve for $x$.

• Substitute $x$ back to find $p(x)$ and $R(x)$.

Additional info: This is a standard quadratic optimization problem.

Example 9: Maximizing Profit

Problem: $p(x) = 55 - 2x$, $C(x) = 21x + 98$. Find the maximum profit.

• Revenue: $R(x) = x(55 - 2x) = 55x - 2x^2$

• Profit: $P(x) = R(x) - C(x) = 55x - 2x^2 - 21x - 98 = 34x - 2x^2 - 98$

• Set derivative to zero: $P'(x) = 34 - 4x = 0 \implies x = 8.5$

• Maximum profit: $P(8.5) = 34(8.5) - 2(8.5)^2 - 98$

Example 10: Economic Order Quantity (EOQ)

Problem: Annual demand $D = 50,000$, holding cost $H = $4/case/year, order cost$S = $50/order. Find the optimal order size.

• EOQ formula: $Q^* = \sqrt{\frac{2DS}{H}}$

• Substitute values: $Q^* = \sqrt{\frac{2 \times 50000 \times 50}{4}} = \sqrt{1,250,000} \approx 1118$ cases

Example 11: Maximizing Revenue with Linear Demand

Problem: Ticket price increases from $50$(attendance $4000$) to $52$(attendance $3800$). Find the price that maximizes revenue.

• Find demand equation: $p = m x + b$ using two points.

• Revenue: $R(x) = x p(x)$

• Optimize $R(x)$ as above.

Example 12: Maximizing Revenue and Profit for Funnel Cakes

Problem: At $p = 5.90$, $x = 2750$; at $p = 6.20$, $x = 2000$. (a) Find price maximizing revenue. (b) If $C(x) = 2500 + 0.6x$, find price maximizing profit.

• Find linear demand: $p(x) = m x + b$

• Revenue: $R(x) = x p(x)$, maximize as before.

• Profit: $P(x) = R(x) - C(x)$, maximize as before.

Summary Table: Key Optimization Steps

Step	Description
1. Define Variables	Assign variables to unknowns in the problem.
2. Objective Function	Write the function to be maximized or minimized.
3. Constraint Equation	Express the problem's limitations as an equation.
4. Substitute	Reduce the objective function to one variable using the constraint.
5. Differentiate	Find the derivative of the objective function.
6. Solve	Set the derivative to zero and solve for the variable.
7. Verify	Check if the solution is a maximum or minimum and answer the question.

Additional Info

• Optimization problems are central to business calculus, especially in maximizing profit, minimizing cost, and efficient resource allocation.

• Marginal analysis (using derivatives) is a key tool for decision-making in economics and business.

• Many real-world problems can be modeled and solved using the steps outlined above.