Question

An automobile manufacturer has determined that its annual labor and equipment cost (in millions of dollars) can be modeled by \(C(x, y)=2 x^{2}+3 y^{2}-15 x-20 y+4 x y+39\) where \(x\) is the amount spent per year on labor and \(y\) is the amount spent per year on equipment (both in millions of dollars). Find the values of \(x\) and \(y\) that minimize the annual labor and equipment cost. What is this cost?

Short answer

The values that minimize the annual labor and equipment cost are \(x = 3/4\) million dollars and \(y = 3\) million dollars. The minimum cost is \(31.5\) million dollars.

Step-by-step solution

Find the Partial Derivatives

We start by finding the partial derivatives of \(C\) with respect to \(x\) and \(y\). The partial derivative of \(C\) with respect to \(x\) is \[C_x = 4x - 15 + 4y,\] and the partial derivative of \(C\) with respect to \(y\) is \[C_y = 6y - 20 + 4x.\]

Find the Critical Points

We then set these partial derivatives equal to zero and solve for the variables \(x\) and \(y\). From \(C_x = 0\) we get \(x = 15/4 - y\) and from \(C_y = 0\) we get \(y = 20/6 - 2x/3 = 10/3 - 2x/3.\) Substituting the \(x\) value from \(C_x = 0\) into the equation from \(C_y = 0\), we obtain \(y = 10/3 - 2(15/4 - y)/3\), solving this we get \(y = 3.\) Substituting \(y = 3\) into the \(x = 15/4 - y\) we get \(x = 3/4.\)

Minimum cost

Substitute the values of \(x = 3/4\) and \(y = 3\) into the cost function \(C(x,y)\) to get the minimum cost. After the calculation, the minimum cost is \(31.5\) million dollars.

Key concepts

Partial Derivatives

Understanding partial derivatives is crucial when dealing with functions of several variables, as in the optimization of a cost function. In essence, a partial derivative represents how the function changes as one variable changes, while the other variables are held constant. Consider a cost function like the one given, \( C(x, y) = 2x^2 + 3y^2 - 15x - 20y + 4xy + 39 \).

To find the partial derivatives, we differentiate with respect to each variable independently. The partial derivative with respect to \(x\), denoted \(C_x\), gives us \(4x - 15 + 4y\), and with respect to \(y\), denoted \(C_y\), it is \(6y - 20 + 4x\). These derivatives guide us toward the function's behavior concerning each direction — labor and equipment cost in this scenario — and are the first step in optimizing the cost.

Critical Points

Critical points in multivariable calculus are where the first derivatives (in this case, the partial derivatives) are zero or undefined. These points give us potential locations of local maxima, minima, or saddle points. To locate the critical points for our cost function, we set the partial derivatives equal to zero, \(C_x = 0\) and \(C_y = 0\), and solve the resulting system of equations.

For the given cost function, solving \(C_x = 0\) and \(C_y = 0\) provides us with the values of \(x\) and \(y\) that potentially minimize or maximize the function. These points are essential for businesses or individuals looking to optimize costs since they indicate where a slight change in the input variables will not substantially change the cost, at least in the immediate vicinity of those critical points.

Cost Function Optimization

Optimizing a cost function often involves minimizing expenses or maximizing efficiency. For a function like \( C(x, y) \), we seek the combination of \(x\) and \(y\) that will give the lowest possible value, representing the most efficient allocation of resources. After locating the critical points, we perform further tests to confirm whether the points actually correspond to a minimum cost.

In our example, substituting the found critical values of \( x = 3/4 \) and \( y = 3 \) back into the original function gives us the minimum cost of \( 31.5 \) million dollars. By ensuring this point is indeed a minimum through second-derivative test or other analytic methods if necessary, businesses can confidently adjust their spending to achieve this optimized cost.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions of several variables. It helps us understand and analyze complex systems where multiple factors influence the outcome simultaneously. In such systems, calculating partial derivatives, identifying critical points, and optimizing functions become multidimensional problems requiring careful consideration of how each variable contributes to the whole.

For example, with a cost function that depends on both labor and equipment costs, multivariable calculus enables us to understand how changing each, or both, impacts the total cost. This mathematical discipline is indispensable in economics, engineering, physical sciences, and several other fields where optimization is key.