University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

242

Pages:

3
Author:

Herbosambx

Question 1 An objective function and a system of linear inequalities representing constraints are given: 𝑧 = 2 𝑥 + 3 𝑦 z=2x+3y 𝑥 ≥ 0 , 𝑦 ≥ 0 x≥0,y≥0 2 𝑥 + 𝑦 ≤ 8 2x+y≤8 2 𝑥 + 3 𝑦 ≤ 12 2x+3y≤12 a) Graph the system of inequalities representing the constraints. b) Find the value of the objective function at each corner of the graphed region. c) Use the values in part (b) to determine the maximum value of the objective function and the values of 𝑥 x and 𝑦 y for which the maximum occurs. Answer 1 Steps: Constraints: 𝑥 ≥ 0 x≥0 (1) 𝑦 ≥ 0 y≥0 (2) 2 𝑥 + 𝑦 ≤ 8 2x+y≤8 (3) 2 𝑥 + 3 𝑦 ≤ 12 2x+3y≤12 (4) a) Graph of given inequalities is provided. b) The value at the corner points is: At point 𝐴 ( 0 , 0 ) A(0,0): 𝑧 = 2 ( 0 ) + 3 ( 0 ) = 0 z=2(0)+3(0)=0 At point 𝐵 ( 0 , 4 ) B(0,4): 𝑧 = 2 ( 0 ) + 3 ( 4 ) = 12 z=2(0)+3(4)=12 At point 𝐶 ( 3 , 2 ) C(3,2): 𝑧 = 2 ( 3 ) + 3 ( 2 ) = 12 z=2(3)+3(2)=12 At point 𝐷 ( 4 , 0 ) D(4,0): 𝑧 = 2 ( 4 ) + 3 ( 0 ) = 8 z=2(4)+3(0)=8 c) The maximum value of 𝑧 z is 12 at the points 𝐵 ( 0 , 4 ) B(0,4) and 𝐶 ( 3 , 2 ) C(3,2). Graph provided. Question 2 Graph the solution set for the system of linear inequalities: 𝑥 ≥ 3 , x≥3, 𝑦 ≥ − 1. y≥−1. Answer 1 Steps: Auxiliary equations are 𝑥 = 3 x=3 and 𝑦 = − 1 y=−1. Plot the lines 𝑥 = 3 x=3 (a vertical line passing through 𝑥 = 3 x=3) and 𝑦 = − 1 y=−1 (a horizontal line passing through 𝑦 = − 1 y=−1). Shade the regions for 𝑥 ≥ 3 x≥3 (right of 𝑥 = 3 x=3) and 𝑦 ≥ − 1 y≥−1 (above 𝑦 = − 1 y=−1). Graph provided. Question 3 Graph the solution set of the inequality or system of inequalities: 𝑥 ≥ 0 , 𝑦 ≥ 0 , x≥0,y≥0, 3 𝑥 + 𝑦 ≤ 9 , 3x+y≤9, 2 𝑥 + 3 𝑦 ≥ 6. 2x+3y≥6. Answer 1 The graph is shown with shaded regions.