PRECALCULUS:GRAPHICAL,...-NASTA ED.
PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter 7.4, Problem 34E
Solution

To calculate: The minimum and minimum value, checked they exists, of the objective function f .

The common region in the graph is the solution of the inequality.

Given Information:

The system of inequality is,

  Objective function:f=15x+25y3x+4y≥60x+8y≥40Constraints:11x+28y≤380x≥0,y≥0

Calculation:

Consider the given system of inequality,

  Objective function:f=15x+25y3x+4y≥60x+8y≥40Constraints:11x+28y≤380x≥0,y≥0

Use the graphing calculator to draw the inequality.

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 7.4, Problem 34E , additional homework tip 1

The polygonal region of feasible solutions has three vertices. To find these, solve each of the following system of boundary equations:

Write the inequality as equation and find the intersecting point.

  3x+4y=60x+8y=4011x+28y=380

Find the intersecting points between the equations.

Substitute x=40−8y in the first equation.

  340−8y+4y=60120−24y+4y=6060=20yy=3

The obtained point is 16,3 .

Substitute x=8y−40 in the third equation.

  1140−8y+28y=380440−88y+28=38060=60yy=1

The obtained point is 32,1 .

Substitute x=60−4y3 in the third equation.

  1160−4y3+28y=380660−44y+84y3=38040y=1140−660y=12

The obtained point is 4,12 .

So, mentioned the point on the graph and shaded the common region.

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 7.4, Problem 34E , additional homework tip 2

Find the value of the objective function at the all vertices.

Substitute 16 for x and 3 for y in the objective function.

  f=1516+253=315

Substitute 32 for x and 1 for y in the objective function.

  f=1532+251=505

Substitute 48 for x and 2 for y in the objective function.

  f=154+2512=360

Therefore, the minimum value is 315 and the maximum value is 505.

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Chapter 7.4, Problem 33E
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Chapter 7.4, Problem 35E

Chapter 7 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

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