Algebra 2
Algebra 2
1st Edition
ISBN: 9780078884825
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 3.4, Problem 10PPS
To determine

To calculate: The coordinates of the feasible region when the provided system of inequalities are graphed. Also evaluate the minimum and maximum value of the function f(x,y)=8x−10y.

  −2≤x≤45≤y≤82x+3y≤26

Expert Solution & Answer
Check Mark

Answer to Problem 10PPS

The coordinates of the feasible region are (−2,8),(−2,5),(4,5),(4,6) and (1,8) . The maximum value of the function is −18 which occurs at (4,5) . The minimum value of the function is −96 which occurs at (−2,8).

  Algebra 2, Chapter 3.4, Problem 10PPS , additional homework tip 1

Explanation of Solution

Given information:

The system of inequalities,

  −2≤x≤45≤y≤82x+3y≤26

Calculation:

Consider the system of inequalities,

  −2≤x≤45≤y≤82x+3y≤26

Plot the inequalities −2≤x≤4, 5≤y≤8 and 2x+3y≤26 on the coordinate plane.

  Algebra 2, Chapter 3.4, Problem 10PPS , additional homework tip 2

The feasible region is the region that is common to all the three inequalities.

Therefore, the region bounded by the vertices (−2,8),(−2,5),(4,5),(4,6) and (1,8) is the feasible region.

Now, substitute each of the vertices in the function f(x,y)=8x−10y to compute the maximum and minimum value of the function.

Value of the function at the vertex (−2,8) ,

  f(−2,8)=8(−2)−10(8)=−16−80=−96

Value of the function at the vertex (−2,5) ,

  f(−2,5)=8(−2)−10(5)=−16−50=−66

Value of the function at the vertex (4,5) ,

  f(4,5)=8(4)−10(5)=32−50=−18

Value of the function at the vertex (4,6) ,

  f(4,6)=8(4)−10(6)=32−60=−28

Value of the function at the vertex (1,8) ,

  f(1,8)=8(1)−10(8)=8−80=−72

Therefore, the maximum value of the function is −18 which occurs at (4,5) . The minimum value of the function is −96 which occurs at (−2,8).

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Chapter 3 Solutions

Algebra 2

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