5 (variant of Strayer Ch. 1 # 10):(a) Draw the constraint set defined by the constraints x + y > 2, x – 2y < 0, y – 2x < 1, andX, y > 0.(b) Argue that the minimum of g(x, y) = x – 3y on this constraint set does not exist, byconsidering points on the line x – 2y = 0 that go off to infinity (but are still in the constraintset).(c) Explain why g< 0 for every point in the constraint set.(d) Based on (c), the maximum of g(x, y) on this constraint set does exist. Use your picture(and some other computations) to find it.

Name: 5 (variant of Strayer Ch. 1 # 10):(a) Draw the constraint set defined
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Description: 5 (variant of Strayer Ch. 1 # 10):(a) Draw the constraint set defined by the constraints x + y > 2, x – 2y 0.(b) Argue that the minimum of g(x, y) = x – 3y on this constraint set does not exist, byconsidering points on the

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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5 (variant of Strayer Ch. 1 # 10):
(a) Draw the constraint set defined by the constraints x + y > 2, x – 2y < 0, y – 2x < 1, and
X, y > 0.
(b) Argue that the minimum of g(x, y) = x – 3y on this constraint set does not exist, by
considering points on the line x – 2y = 0 that go off to infinity (but are still in the constraint
set).
(c) Explain why g< 0 for every point in the constraint set.
(d) Based on (c), the maximum of g(x, y) on this constraint set does exist. Use your picture
(and some other computations) to find it.

Expert Solution

Step 1

By bartleby rules, only first three subparts have been answered

To draw the constraint set defined by

$\begin{array}{rcl} x + y & \geq & 2 \\ x - 2 y & \leq & 0 \\ y - 2 x & \leq & 1 \\ x, y & \geq & 0 \end{array}$

First let us consider the corresponding equation forms of the inequalities.

$x + y = 2$ (i)

This line intersect x-axis at (2,0) and y-axis at (0,2). Also from the inequality, it is seen that $y \geq 2 - x$ . The inequality sign is $\geq$ . therefore, all the points above the line must be considered.

$\begin{array}{rcl} x - 2 y & = & 0 \\ y & = & \frac{x}{2} \end{array}$

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