Problem 18 - Calculus and its applications

Question

Assume that relative maximum and minimum values are absolute maximum and minimum values. Humphrey's Medical Supply finds that its profit, \(P\), in millions of dollars, is given by $$ P(a, n)=-5 a^{2}-3 n^{2}+48 a-4 n+2 a n+290 $$ where \(a\) is the amount spent on advertising, in millions of dollars, and \(n\) is the number of items sold, in thousands. Find the maximum value of \(P\) and the values of \(a\) and \(n\) at which it occurs

Step-by-Step Solution

Answer

The maximum profit is 408 million dollars at \(a = 5\) and \(n = 1\).

Step 1: Determine the Partial Derivatives

To find where the maximum occurs, we first compute the partial derivatives of the profit function \(P(a, n)\). These derivatives are \( \frac{\partial P}{\partial a} \) and \( \frac{\partial P}{\partial n} \). Calculate:\( \frac{\partial P}{\partial a} = -10a + 2n + 48 \) and \( \frac{\partial P}{\partial n} = -6n + 2a - 4 \).

Step 2: Set Partial Derivatives to Zero

To find the critical points, set each of the partial derivatives equal to zero.\(-10a + 2n + 48 = 0 \quad \text{(1)}\)\(-6n + 2a - 4 = 0 \quad \text{(2)}\)

Step 3: Solve the System of Equations

Use equations (1) and (2) to solve for \(a\) and \(n\).From equation (2):\[a = 3n + 2 \quad \text{(3)}\]Substitute equation (3) into equation (1):\(-10(3n + 2) + 2n + 48 = 0\)Simplify to find \(n\):\(-30n - 20 + 2n + 48 = 0\)\(-28n + 28 = 0\)\(\Rightarrow n = 1\)Substitute \(n = 1\) back into equation (3):\( a = 3(1) + 2 = 5\)

Step 4: Evaluate the Profit at Critical Points

Substitute \(a = 5\) and \(n = 1\) back into the original profit function \(P(a, n)\):\[P(5, 1) = -5(5)^2 - 3(1)^2 + 48(5) - 4(1) + 2(5)(1) + 290\]Simplify this:\[P(5, 1) = -125 - 3 + 240 - 4 + 10 + 290 = 408\]

Step 5: Conclude the Maximum Profit and Variables

The maximum profit value calculated from the given function is \(408\) million dollars. This occurs when the advertising expenditure \(a\) is 5 million dollars and the number of items sold \(n\) is 1 thousand items.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially useful in optimization problems involving functions of multiple variables. When you have a function like the profit function, which depends on more than one variable, partial derivatives help you understand how the function changes as each variable changes individually.

In this problem, the profit function is represented as a function of two variables: advertising expenditure (\(a\)) and the number of items sold (\(n\)). The partial derivative with respect to \(a\), denoted \(\frac{\partial P}{\partial a}\), measures the rate of change of the profit while keeping \(n\) constant. Similarly, the partial derivative with respect to \(n\), denoted \(\frac{\partial P}{\partial n}\), measures the rate of change of the profit with respect to \(n\) while keeping \(a\) constant.

• To compute these derivatives, you differentiate the profit equation concerning each variable. It's like peeling away one layer at a time to see how each dimension affects your profit.
• Partial derivatives are crucial for understanding the behavior of multi-variable functions, revealing how small adjustments in one variable can impact the overall function value.

Critical Points

Finding critical points is a key step in optimization problems. Critical points occur where the partial derivatives of a function are zero or it's undefined. For the given profit function, you found the partial derivatives with respect to \(a\) and \(n\).

By setting these derivatives equal to zero, you can find potential points where the profit might reach a maximum or a minimum. Why? Because these points indicate where the surface created by the function \(P(a, n)\) is flat in every direction, like a peak or a valley.

• For a practical scenario, solving the equations \(-10a + 2n + 48 = 0\) and \(-6n + 2a - 4 = 0\) allowed you to pinpoint the values of \(a\) and \(n\) at critical points.
• In this exercise, substituting back the solved values helps confirm if they indeed give the highest possible profit or if other potential critical points exist.

Profit Function Analysis

Once the critical points are found, the analysis doesn't stop there. Evaluating the profit function at these points determines if they actually correspond to a maximum profit. Understanding this analysis helps make informed decisions in business settings, like how much to spend on advertising or how many items to sell.

With the critical points obtained, in this case \(a = 5\) and \(n = 1\), you substitute these back into the original profit function. The result gives the maximum possible value for the profit in millions of dollars.

• This step is crucial because not every critical point results in a maximum profit. Without checking, you might mistake a minimum or a saddle point as the optimal solution.
• Verifying \(P(a, n)\) through substitution at critical points ensures that you've truly found where profit is maximized, not merely at any flat part of the function surface.