Question

In order to treat a certain bacterial infection, a combination of two drugs is being tested. Studies have shown that the duration of the infection in laboratory tests can be modeled by \(D(x, y)=x^{2}+2 y^{2}-18 x-24 y+2 x y+120\) where \(x\) is the dosage in hundreds of milligrams of the first drug and \(y\) is the dosage in hundreds of milligrams of the second drug. Determine the partial derivatives of \(D\) with respect to \(x\) and with respect to \(y\). Find the amount of each drug necessary to minimize the duration of the infection.

Short answer

The amount of each drug necessary to minimize the duration of the infection is 600mg for the first drug and 300mg for the second drug.

Step-by-step solution

Define the function

The function is defined as \(D(x, y)=x^{2}+2 y^{2}-18 x-24 y+2 x y+120\), which represents the duration of the infection.

Find the partial derivatives

Find the first partial derivatives of the function \(D\) with respect to both \(x\) and \(y\). The partial derivative with respect to \(x\) is computed as \(\frac{{\partial D}}{{\partial x}} = 2x - 18 + 2y\), and with respect to \(y\) it is \(\frac{{\partial D}}{{\partial y}} = 4y - 24 + 2x\).

Set the partial derivatives to zero

To find the amount of each drug that will minimize the duration, set the partial derivatives to zero and solve for \(x\) and \(y\). This yields the following system of equations: \(2x - 18 + 2y = 0\) (1), and \(4y - 24 + 2x = 0\) (2).

Solve the system of equations

Solving the system of equations gets \(x = 6\) and \(y = 3\). This means that, to minimize the duration of the infection, the dosage should be 600mg for the first drug and 300mg for the second drug.

Verify the minimum

To make sure that these values actually give the minimum duration, we should confirm that the second partial derivatives are positive and that the determinant of the Hessian matrix is positive. This step is left to be calculated.

Key concepts

Multivariable Calculus

Multivariable calculus is an expansion of calculus into the realm of more than one variable. Unlike single-variable calculus that deals with functions of one variable, multivariable calculus handles functions of two or more variables. It is particularly useful in studying systems where multiple factors influence the outcomes and is fundamental in many areas of science and engineering.

In our exercise example, we have a function, \(D(x, y)\), which depends on two variables: \(x\) and \(y\). Each represents the dosage of different drugs in hundreds of milligrams, and the function itself represents the duration of a bacterial infection. The goal is typically to understand how changes in multiple inputs can simultaneously affect an outcome, which in this case, is the duration of the infection.

The process involves taking partial derivatives which tell us how the function changes as one variable changes while the others are held constant. For optimization problems, such as finding the minimum duration of the infection by adjusting the dosages of drugs, multivariable calculus is essential.

Optimization Problem

An optimization problem in calculus is one where you seek the maximum or minimum value of a function given some constraints. In multivariable calculus, optimization problems can be more complex due to the presence of multiple variables and potentially several local minima and maxima.

In our exercise, the essential question is: What dosages of two drugs minimize the duration of an infection? To solve this, we first define our objective function, \(D(x, y)\), representing the duration of the infection. We then compute the first partial derivatives with respect to both variables, which gives us the rate of change of the infection duration with changes in drug dosages.

We find the critical points by setting these partial derivatives to zero because at these points, the rate of change of the duration is zero in the direction of any variable, indicating a local extremum - a possible minimum or maximum. After solving these equations, we obtain the dosage that potentially minimizes the infection duration. However, confirmation is still needed that these points provide a minimum, not a maximum or saddle point, which brings us to analyzing second partial derivatives.

Second Partial Derivatives

Second partial derivatives play a crucial role in multivariable calculus, especially in the context of optimization. After finding the critical points, second partial derivatives help us determine whether each point is a local maximum, minimum, or a saddle point. For a function of two variables, \(D(x, y)\), we consider the second partial derivatives \(D_{xx}\), \(D_{yy}\), and the mixed derivative \(D_{xy}\)=\(D_{yx}\).

The Hessian matrix is constructed from these second partial derivatives and provides the necessary criteria for testing whether a critical point is a relative minimum or maximum. It is a square matrix of all the second-order partial derivatives.

To confirm that we have a minimum in our exercise's context, we must check two conditions: the second partial derivatives at the critical point are positive (implying the function is locally convex), and the determinant of the Hessian is positive. If these conditions are met for the critical points obtained earlier (600mg of the first drug and 300mg of the second drug), we have successfully minimized the duration of the infection.