Problem 45 - College Algebra and Calculus: An Applied Approach

Question

An animal shelter buys two different brands of dog food. The number of dogs that can be fed from \(x\) pounds of the first brand and \(y\) pounds of the second brand is given by the model \(D(x, y)=-x^{2}+52 x-y^{2}+44 y+256\) (a) The shelter orders 100 pounds of \(\operatorname{dog}\) food. Use Lagrange multipliers to find the number of pounds of each brand of dog food that should be in the order so that the maximum number of dogs can be fed. (b) What is the maximum number of dogs that can be fed?

Step-by-Step Solution

Answer

The shelter should buy 52 pounds of the first brand and 48 pounds of the second brand to maximize the number of dogs fed, and this will feed a maximum of 2844 dogs.

Step 1: Formulating the Lagrange System

In this situation, the Lagrange function will be the vehicle for a system of equations. The function is formed by \(D(x, y) - \lambda (x + y - 100)\), where \(\lambda\) is the Lagrange multiplier. Thus, the Lagrange system is formed by the gradient of the Lagrange function to zero, giving three equations: \n1. \(-2x + 52 - \lambda = 0\)\n2. \(-2y + 44 - \lambda = 0\)\n3. \(x + y - 100 = 0\)

Step 2: Solving the System of Equations

Now that we have formulated our system of equations, we can solve them. Solve the first two equations for \(\lambda\), we get \(\lambda = 52 - 2x\) from the first equation and \(\lambda = 44 - 2y\) from the second equation. We now set these two equations equal to each other and solve for y giving us \(y = x - 4\).\nSubstitute \(y = x - 4\) into the third equation \(x + y - 100 = 0\) to get \(x = 52\) and \(y = 48\). So, to feed the maximum number of dogs, the shelter should buy 52 pounds of the first brand and 48 pounds of the second brand of dog food.

Step 3: Finding the Maximum Number of Dogs

To find the maximum number of dogs that can be fed, given the pounds for the first and second brand of dog food, we substitute \(x = 52\) and \(y = 48\) into the dog feeding model \(D(x, y) = -x^{2}+52x-y^{2}+44y+256.\) This gives us \(D(52, 48) = 2844.\)

Key Concepts

College Algebra

College algebra is a key foundation in mathematics that forms the basis for many theoretical and applied disciplines. One of the fundamental skills you develop through college algebra is the ability to manipulate algebraic expressions and solve equations. While working on an optimization problem involving an animal shelter's purchase of dog food to feed the maximum number of dogs, you encounter a practical application of algebra. Here, we use algebraic methods to establish relationships between quantities of dog food and the number of dogs that can be fed.

Formulating an equation like the given model, \(D(x, y)=-x^{2}+52x-y^{2}+44y+256\), requires an understanding of how variables interact within an equation. This model represents a quadratic function in two variables, where the coefficients of \(x\) and \(y\) have real-world implications, corresponding to the efficiency of each brand in feeding dogs. Algebra allows us to engage with this model and explore its implications under restrictions such as a fixed total amount of dog food.

Systems of Equations

In multivariable optimization problems like this one involving the animal shelter, a system of equations emerges as a powerful tool. Systems of equations are sets of equations with multiple variables, where each equation represents a relationship between those variables. Instead of looking for a single solution, we're interested in the set of all solutions that satisfy every equation simultaneously.

For example, the Langrange system in this exercise gives us three equations: \(-2x + 52 - \(\lambda\) = 0\), \(-2y + 44 - \(\lambda\) = 0\), and \(x + y - 100 = 0\). These equations link the quantities of dog food and an auxiliary variable, \(\lambda\), which is introduced to handle the constraint that the total pounds of dog food must be 100. By using algebraic techniques to manipulate and solve these equations, we can find that the optimal solution is to purchase 52 pounds of the first brand and 48 pounds of the second. The use of systems of equations here illustrates how multiple computational strands can be woven together to achieve a desired outcome.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables, like our dog feeding model \(D(x, y)\). In such problems, optimization often involves finding the maximum or minimum values of a function subject to constraints. This is where Lagrange multipliers come into play as a technique for constrained optimization.

The idea is to introduce an auxiliary variable, the Lagrange multiplier \(\lambda\), that scales the gradient of the constraint function to make it equal to the gradient of the objective function. In simple terms, we're balancing the rate at which our function \(D(x, y)\) changes with the rates at which the constraints are changing. This balance point gives us potential candidates for maximum or minimum values – in this case, the maximum number of dogs that can be fed, which is 2844 when 52 pounds of the first brand and 48 pounds of the second brand are purchased. Multivariable calculus is crucial for solving complex real-world problems where multiple factors come into play simultaneously.