Problem 15 - Calculus and its applications

Question

Assume that relative maximum and minimum values are absolute maximum and minimum values. Safe Shades produces two kinds of sunglasses; one kind sells for \(\$ 17,\) and the other for \(\$ 21\). The total revenue in thousands of dollars from the sale of \(x\) thousand sunglasses at \(\$ 17\) each and \(y\) thousand at \(\$ 21\) each is given by $$ R(x, y)=17 x+21 y $$ The company determines that the total cost, in thousands of dollars, of producing \(x\) thousand of the \(\$ 17\) sun glasses and \(y\) thousand of the \(\$ 21\) sunglasses is given by $$ C(x, y)=4 x^{2}-4 x y+2 y^{2}-11 x+25 y-3 $$ How many of each type of sunglasses must be produced and sold to maximize profit?

Step-by-Step Solution

Answer

Produce and sell 6,000 sunglasses at $17 and 5,000 sunglasses at $21 for maximum profit.

Step 1: Define Profit Function

The profit function is given by subtracting the cost function from the revenue function. We have:\[ P(x, y) = R(x, y) - C(x, y) = (17x + 21y) - (4x^2 - 4xy + 2y^2 - 11x + 25y - 3) \]Simplify this to: \[ P(x, y) = -4x^2 + 4xy - 2y^2 + 28x - 4y + 3 \]

Step 2: Find Critical Points

To find the critical points, we need to take the first partial derivatives of the profit function with respect to \(x\) and \(y\) and set them equal to zero.Partial derivative with respect to \(x\): \[ \frac{\partial P}{\partial x} = -8x + 4y + 28 \]Setting it to zero: \[ -8x + 4y + 28 = 0 \] (Equation 1)Partial derivative with respect to \(y\): \[ \frac{\partial P}{\partial y} = 4x - 4y - 4 \]Setting it to zero: \[ 4x - 4y - 4 = 0 \] (Equation 2)

Step 3: Solve the System of Equations

Using Equation 1 and Equation 2, solve the system of equations:From Equation 2: \[ 4x - 4y = 4 \]This implies: \[ x = y + 1 \]Substitute \(x = y + 1\) in Equation 1:\[ -8(y + 1) + 4y + 28 = 0 \]Simplify this:\[ -8y - 8 + 4y + 28 = 0 \]\[ -4y + 20 = 0 \] Solve for \(y\):\[ y = 5 \]Substitute \(y = 5\) back into \(x = y + 1\): \[ x = 5 + 1 = 6 \]

Step 4: Verify the Critical Point

Check if the critical point \((x, y) = (6, 5)\) actually provides a maximum by using the second partial derivative test. Compute the relevant second derivatives:\( \frac{\partial^2 P}{\partial x^2} = -8 \)\( \frac{\partial^2 P}{\partial y^2} = -4 \)\( \frac{\partial^2 P}{\partial x \partial y} = 4 \)Check the determinant of the Hessian matrix:\[ D = \frac{\partial^2 P}{\partial x^2} \frac{\partial^2 P}{\partial y^2} - \left( \frac{\partial^2 P}{\partial x \partial y} \right)^2 \]\[ D = (-8)(-4) - (4)^2 = 32 - 16 = 16 \]Since \(D > 0\) and \( \frac{\partial^2 P}{\partial x^2} < 0 \), the point is a local maximum.

Step 5: State the Conclusion

The maximum profit occurs when producing and selling 6 thousand of the \(\\( 17\) sunglasses and 5 thousand of the \(\\) 21\) sunglasses.

Key Concepts

Partial Derivatives

Partial derivatives are a key concept in multivariable calculus, crucial for finding how a function varies with respect to one variable while keeping the others constant. In the context of profit maximization, partial derivatives allow us to observe how changes in the production of sunglasses of different prices impact the total profit.

To compute a partial derivative, we treat all other variables as constants and differentiate concerning one variable at a time. For example, for the profit function \( P(x, y) = -4x^2 + 4xy - 2y^2 + 28x - 4y + 3 \), the partial derivative with respect to \( x \), \( \frac{\partial P}{\partial x} \), considers \( y \) as a constant. Likewise, \( \frac{\partial P}{\partial y} \) treats \( x \) as a constant.

• \( \frac{\partial P}{\partial x} = -8x + 4y + 28 \)
• \( \frac{\partial P}{\partial y} = 4x - 4y - 4 \)

Partial derivatives help locate points where changes briefly "stall" (where the derivatives equal zero), indicating potential maximum, minimum, or saddle points.

Critical Points

Critical points occur where the first partial derivatives of a function are zero. They are candidate points for local maxima, minima, or saddle points, which are vital when determining optimal production or sales levels.

To find the critical points for our profit function, we set both \( \frac{\partial P}{\partial x} \) and \( \frac{\partial P}{\partial y} \) to zero, creating a system of equations:

• \( -8x + 4y + 28 = 0 \)
• \( 4x - 4y - 4 = 0 \)

Solving this system, substitute \( x = y + 1 \) into the first equation, simplifying and solving for \( y \), yielding \( y = 5 \). Then, substitute back to find \( x \), obtaining \( x = 6 \).

Here, \((x, y) = (6, 5)\) represents a critical point, indicating a potential local maximum based on partial derivative calculation.

Second Partial Derivative Test

The second partial derivative test helps determine the nature of critical points, telling us whether the point is a local maximum, minimum, or saddle point. For our profit function, it requires calculating second-order partial derivatives and the Hessian determinant:

• \( \frac{\partial^2 P}{\partial x^2} = -8 \)
• \( \frac{\partial^2 P}{\partial y^2} = -4 \)
• \( \frac{\partial^2 P}{\partial x \partial y} = 4 \)

To complete the test, calculate the Hessian determinant \( D \):
\[ D = \frac{\partial^2 P}{\partial x^2} \frac{\partial^2 P}{\partial y^2} - \left( \frac{\partial^2 P}{\partial x \partial y} \right)^2 \]
\[ D = (-8)(-4) - (4)^2 = 32 - 16 = 16 \]
Since \( D > 0 \) and \( \frac{\partial^2 P}{\partial x^2} < 0 \), the critical point \((6, 5)\) is confirmed to be a local maximum.

This conclusion is essential for profit maximization, confirming that producing 6,000 of the \(\\(17\) sunglasses and 5,000 of the \(\\)21\) sunglasses optimizes the company's profit. This method ensures businesses efficiently utilize resources to maximize returns.