91Ó°ÊÓ

Profit maximization is a fundamental goal for most companies. To achieve this, a company needs to find the production levels that yield the highest possible profit. Profit is calculated as the difference between revenue and cost. In this exercise, the company sells two products: I and II. The goal is to determine the quantities of each product, denoted as \( x \) and \( y \), that will maximize the company's profit. The revenue function, which depends on the selling prices and quantities of the products, is defined as \( R = 10x + 9y \). The cost function, accounting for production costs and additional terms, is given by: \( C = 400 + 2x + 3y + 0.01(3x^2 + xy + 3y^2) \). By defining a profit function \( P \) as \( R - C \), we simplify to find: \( P = 8x + 6y - 400 - 0.01(3x^2 + xy + 3y^2) \). The task now is to find the specific values of \( x \) and \( y \) that maximize this profit function.

Partial derivatives are used to measure how a function changes as its variables change. When considering functions of multiple variables, partial derivatives are essential for optimization. In this exercise, we need to consider \( P(x, y) \), the profit function of \( x \) and \( y \). The first partial derivatives of \( P \) with respect to \( x \) and \( y \) are: \( \frac{\partial P}{\partial x} = 8 - 0.01(6x + y) \) and \( \frac{\partial P}{\partial y} = 6 - 0.01(x + 6y) \). To find the critical points where profit might be maximized, we set these derivatives to zero because at these points, the slope of the function is zero, indicating a potential maximum or minimum. This results in the system of equations: \( 8 - 0.01(6x + y) = 0 \) and \( 6 - 0.01(x + 6y) = 0 \). Solving this system gives us the values of \( x \) and \( y \) at which the profit function's rate of change is zero.

After finding the critical points using the first partial derivatives, it is crucial to determine whether these points are maxima, minima, or saddle points. The second derivative test helps with this, involving the second partial derivatives of the function. For our profit function \( P(x, y) \), we compute: \( \frac{\partial^2 P}{\partial x^2} = -0.06 \), \( \frac{\partial^2 P}{\partial y^2} = -0.06 \), and \( \frac{\partial^2 P}{\partial x \partial y} = -0.01 \). The test uses the Hessian determinant \( H \), calculated as: \[ H = \left( \frac{\partial^2 P}{\partial x^2} \right) \left( \frac{\partial^2 P}{\partial y^2} \right) - \left( \frac{\partial^2 P}{\partial x \partial y} \right)^2 \]. Substituting in our second partial derivatives, \( H = (-0.06)(-0.06) - (-0.01)^2 = 0.0036 - 0.0001 = 0.0035 \), which is greater than zero. Additionally, \( \frac{\partial^2 P}{\partial x^2} < 0 \) and \( \frac{\partial^2 P}{\partial y^2} < 0 \), indicating that the critical points are local maxima.

A system of equations involves multiple equations that we solve simultaneously to find the values of variables that satisfy all the equations. In the context of this exercise, after setting the partial derivatives to zero, we get two equations: \( 8 - 0.01(6x + y) = 0 \) and \( 6 - 0.01(x + 6y) = 0 \). We solve this system step-by-step.
First, simplify the equations: \( 800 = 6x + y \) and \( 600 = x + 6y \). To find \( x \) and \( y \), rearrange the second equation to isolate \( x \): \( x = 600 - 6y \). Substitute this into the first equation:
\[ 800 = 6(600 - 6y) + y \]
Simplify and solve: \( 800 = 3600 - 36y + y \) leading to \( 800 = 3600 - 35y \) and eventually \( 35y = 2800 \). Solving this, we get \( y = 80 \). Substitute \( y \) back into the rearranged second equation to find \( x \): \( x = 600 - 480 = 120 \). Thus, the profit-maximizing quantities are \( x = 120 \) and \( y = 80 \).