Algebra 2
Algebra 2
1st Edition
ISBN: 9780078884825
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 3.5, Problem 36STP
To determine

To calculate: The maximum and minimum values of the function f(x,y)=y−4x in the feasible region represented by given coordinates.

Expert Solution & Answer
Check Mark

Answer to Problem 36STP

The maximum value of the function is 14 at (−3,2) and minimum value of the function is −23 at (6,1) .

Explanation of Solution

Given information:

The vertices of the feasible region (−3,2),(1,3),(6,1) and (2,−2) .

Formula used:

To determine the maximum and minimum value of the function, substitute the coordinates of the feasible region in the function and determine the maximum and the minimum value.

Calculation:

Consider the provided vertices of the feasible region (−3,2),(1,3),(6,1) and (2,−2) .

The feasible region represented by the given vertices is drawn in the following graph,

  Algebra 2, Chapter 3.5, Problem 36STP

Recall that to determine the maximum and minimum value of the function, substitute the coordinates of the feasible region in the function and determine the maximum and the minimum value.

So, substitute (−3,2),(1,3),(6,1) and (2,−2) in the function f(x,y)=y−4x and evaluate the maximum and minimum value as,

  (x,y)y−4xf(x,y)(−3,2)2−4(−3)14(1,3)3−4(1)−1(6,1)1−4(6)−23(2,−2)−2−4(2)−10

From the above table, it is observed that the maximum value of the function is 14 at (−3,2) and minimum value of the function is −23 at (6,1) .

Thus, the maximum value of the function is 14 at (−3,2) and minimum value of the function is −23 at (6,1) .

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Chapter 3.5, Problem 35STP
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Chapter 3.5, Problem 37STP

Chapter 3 Solutions

Algebra 2

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