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Magazine Chapter 6: Problem 2
Find the maximum value of \(f\) subject to the given constraint. $$f(x, y)=2 x y ; \quad 4 x+y=16$$
Short Answer
Expert verified
The maximum value of f is 32.
Step by step solution
01
Identify the Objective Function
The objective function given is the function we want to maximize: \[ f(x, y) = 2xy \]
02
Identify the Constraint
The constraint is given by the equation: \[ 4x + y = 16 \]
03
Solve the Constraint for One Variable
Solve the constraint equation for one variable in terms of the other. Here, solve for y in terms of x: \[ y = 16 - 4x \]
04
Substitute Constraint into Objective Function
Substitute the expression for y from the constraint into the objective function: \[ f(x, y) = 2x(16 - 4x) = 32x - 8x^2 \]
05
Differentiate the New Function
Differentiate the function obtained in the previous step with respect to x and set the derivative equal to zero to find the critical points: \[ f'(x) = 32 - 16x \] Setting the derivative equal to zero gives the equation: \[ 32 - 16x = 0 \]
06
Solve for x
Solving the equation from the previous step yields: \[ x = 2 \]
07
Substitute x Back into the Constraint
Use the value of x found to solve for y: \[ y = 16 - 4(2) = 8 \]
08
Calculate the Maximum Value of f
Substitute the values of x and y back into the objective function to find the maximum value of f: \[ f(2, 8) = 2(2)(8) = 32 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Objective Function
In this exercise, the objective function is the equation we want to maximize. Here, our objective function is given as: \[ f(x, y) = 2xy \]This function tells us how x and y interact to produce a value for f. In optimization problems, the objective function can represent various things, such as profit, area, or any other measurable quantity you want to maximize or minimize.
Constraint Equation
A constraint equation is an additional condition that the solution must satisfy. For our problem, the constraint is given by: \[ 4x + y = 16 \]Constraints limit the values that x and y can take. They are often represented as equations or inequalities. To proceed, solve the constraint for one variable in terms of the other. Here, we solve for y: \[ y = 16 - 4x \]
Critical Points
Critical points are where the first derivative of the function equals zero or is undefined. These points help us locate the maximum or minimum values of the function. After substituting the constraint into the objective function, we differentiate: \[ f(x, y) = 2x(16 - 4x) = 32x - 8x^2 \]Differentiating: \[ f'(x) = 32 - 16x \]Setting this to zero and solving: \[ 32 - 16x = 0 \] \[ x = 2 \]
Differentiation
Differentiation is the process of finding the derivative of a function. It tells us the rate of change of the function with respect to one of its variables. In this problem, we differentiate the objective function with respect to x to find where it increases or decreases. Once substituted correctly, we get a new function that we differentiate: \[ f(x) = 32x - 8x^2 \]The derivative is: \[ f'(x) = 32 - 16x \]Setting the derivative to zero helps us find the critical points for maximizing or minimizing the function.
Maximum Value
After finding the critical points, we substitute them back into the original constraint to find the corresponding y-values. Here, substituting x = 2 into the constraint: \[ y = 16 - 4(2) = 8 \]Finally, we substitute x and y into the objective function to find the maximum value: \[ f(2, 8) = 2(2)(8) = 32 \]Thus, the maximum value of the function is 32.
