 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/5/b/iii/solution

After removing the two redundant inequalities, the feasible region is

y_{1} \geq 0, y_{2} \geq max {0, y_{1} - 1, 3 - \frac{3}{4} y_{1}, 6 - 3 y_{1}} .

Its lower boundary consists of three line segments or rays:

y_{2} = 6 - 3 y_{1}

for

0 \leq y_{1} \leq 4/3

, then

y_{2} = 3 - 3 y_{1} /4

for

4/3 \leq y_{1} \leq 16/7

, then

y_{2} = y_{1} - 1

for

y_{1} \geq 16/7

. The feasible region is everything on or above this boundary in the first quadrant.

The boundary extreme points are

(0, 6), (4/3, 2), (16/7, 9/7) .

Along its three successive pieces, the objective function

2 y_{1} + y_{2}

has respective slopes

- 1

5/4

and

3

y_{1}

increases. Thus it decreases to the middle extreme point and increases thereafter. For fixed

y_{1}

, moving above the boundary only increases the objective function. Hence

y^{*} = (4/3, 2), min (2 y_{1} + y_{2}) = \frac{14}{3} .

As an algebraic check of the graphical answer, the active inequalities give

2 y_{1} + y_{2} = \frac{1}{9} (3 y_{1} + 4 y_{2}) + \frac{5}{9} (3 y_{1} + y_{2}) \geq \frac{12}{9} + \frac{30}{9} = \frac{14}{3},

with equality at the stated extreme point.

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