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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.5.5a

Suppose the objective function P= xy is subject to the constraint 10x + y = 100, where x and y are real numbers.


a. Eliminate the variable y from the objective function so that P is expressed as a function of one variable x.

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1
Start with the constraint equation: 10x + y = 100.
Solve the constraint equation for y in terms of x: y = 100 - 10x.
Substitute the expression for y from the constraint into the objective function P = xy.
This substitution gives P = x(100 - 10x).
Simplify the expression to express P as a function of x: P(x) = 100x - 10x^2.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Objective Function

An objective function is a mathematical expression that defines a quantity to be maximized or minimized, often subject to certain constraints. In this case, the objective function P = xy represents a product of two variables, x and y, which we aim to optimize under the given constraint.
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Constraints

Constraints are conditions or limitations placed on the variables of an optimization problem. Here, the constraint 10x + y = 100 restricts the values that x and y can take, ensuring that any solution must satisfy this linear equation while optimizing the objective function.
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Substitution Method

The substitution method involves replacing one variable in an equation with an expression derived from another equation. In this problem, we will use the constraint to express y in terms of x, allowing us to rewrite the objective function P solely in terms of x, simplifying the optimization process.
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