Practice 1 Solutions

Given the graph, find the minimum and maximum values using the objective function.

$f (x, y) = \frac{1}{2} x + 4 y$

$f (2, 0) = \frac{1}{2} (2) + 4 (0) f (2, 0) = 1 minimum f (4, 0) = \frac{1}{2} (4) + 4 (0) f (4, 0) = 2 f (0, 5) = \frac{1}{2} (0) + 4 (5) f (0, 5) = 20 f (1, 6) = \frac{1}{2} (1) + 4 (6) f (1, 6) = 24.5 maximum$

This problem has one minimum at (2, 0) and one maximum at (1, 6).

$f (x, y) = 5 y - x$

$f (0, 0) = 5 (0) - (0) f (0, 0) = 0 f (3, 3) = 5 (3) - (3) f (3, 3) = 12 f (1, 4) = 5 (4) - (1) f (1, 4) = 19 maximum f (4, 1) = 5 (1) - (4) f (4, 1) = 1 minimum$

This problem has one minimum at (4, 1) and one maximum at (1, 4).

Write a system of inequalities given the graph. Then use the objective function to find the minimum and maximum.

$f (x, y) = 6 x + 5 y$

$line a : m = - \frac{1}{2}, y \geq (shading above) y - 0 \geq - \frac{1}{2} (x - 7) y \geq - \frac{1}{2} x - \frac{7}{2} line b : m = - 3, b = 6, y \geq (shading above) y \geq - 3 x + 6$

$f (1, 3) = 6 (1) + 5 (3) f (1, 3) = 21 minimum f (7, 0) = 6 (7) + 5 (0) f (7, 0) = 42 f (0, 6) = 6 (0) + 5 (6) f (0, 6) = 30$

Note

The inequality with a fractional y‑intercept can be found by using the points (1, 3) and (7, 0).

$y \geq - 3 x + 6 y \geq - \frac{1}{2} x + \frac{7}{2} y \geq 0 x \geq 0 f (1, 3) = 21 minimum$

The minimum is at (1, 3). There is no maximum because this is unbounded.

Note

Q: Why is there no maximum for this graph?
A: This graph does not have an enclosed polygon, so this system is unbounded and therefore does not have a max.

$f (x, y) = x - \frac{3}{4} y$

$line w : m = - \frac{1}{4}, b = 0, y \geq (shading above) y \geq - \frac{1}{4} x line x : m = - \frac{1}{4}, b = 4, y \leq (shading below) y \leq - \frac{1}{4} x + 4 line y : m = undefined, x \geq 0 (shading right) x \geq 0 line z : m = 4, y \geq (shading above) y - 3 \geq 4 (x - 4) y - 3 \geq 4 x - 16 y \geq 4 x - 13$

$f (0, 0) = (0) - \frac{3}{4} (0) f (0, 0) = 0 f (0, 4) = 0 - \frac{3}{4} (4) f (0, 4) = - 3 minimum f (3, - 1) = 3 - \frac{3}{4} (- 1) f (3, - 1) = 3.75 maximum f (4, 3) = 4 - \frac{3}{4} (3) f (4, 3) = 1.75$

Note

This can be found by m = 4,
(4, 3). Then use the point slope formula.

$y \geq - \frac{1}{3} x x \geq 0 y \geq - \frac{1}{4} x + 4 y \geq 4 x - 13$

This problem has one minimum at (0, 4) and one maximum at (3, −1).

Note

You can check the inequalities using technology. The graph should match the one provided.

Q: How many inequalities are needed for the bounded region? Explain.
A: 4, because there are 4 graphed lines on the coordinate plane.

Note

Problems 5–10
You can use technology to graph these problems; however, you also need to be able to transfer the information onto a coordinate plane on paper.

Graph the system of inequalities. Name all of the vertices and evaluate using the objective function for the minimum and maximum values.

$x + 2 y \leq 12 x + y \geq 5 x + 3 y \leq 15 x \geq 0 y \geq 0 f (x, y) = 4.5 y - 3 x$

$f (5, 0) = 4.5 (0) - 3 (5) f (5, 0) = - 15 f (0, 5) = 4.5 (5) - 3 (0) f (0, 5) = 22.5 maximum f (12, 0) = 4.5 (0) - 3 (12) f (12, 0) = - 36 minimum f (6, 3) = 4.5 (3) - 3 (6) f (6, 3) = - 4.5$

This problem has one minimum at (12, 0) and one maximum at (0, 5).

$y \leq - x + 4 y \leq - \frac{1}{3} x + 2 x \geq 0 y \geq 0 f (x, y) = 3 x + 2 y$

$f (0, 0) = 3 (0) + 2 (0) f (0, 0) = 0 minimum f (3, 1) = 3 (3) + 2 (1) f (3, 1) = 11 f (0, 2) = 3 (0) + 2 (2) f (0, 2) = 4 f (4, 0) = 3 (4) + 2 (0) f (4, 0) = 12 maximum$

This problem has one minimum at (0, 0) and one maximum at (4, 0).

$y \leq \frac{1}{2} x + 2 y \leq - x + 8 x \geq 2 y \geq 1 f (x, y) = 2 x - 3 y$

$f (2, 1) = 2 (2) - 3 (1) f (2, 1) = 1 f (4, 4) = 2 (4) - 3 (4) f (4, 4) = - 4 f (2, 3) = 2 (2) - 3 (3) f (2, 3) = - 5 minimum f (7, 1) = 2 (7) - 3 (1) f (7, 1) = 11 maximum$

This problem has one minimum at (2, 3) and one maximum at (7, 1).

$y \geq - 3 x + 7 x + 2 y \geq 9 x \geq 0 y \geq 0 f (x, y) = 4 x - y$

$f (1, 4) = 4 (1) - 4 f (1, 4) = 0 f (9, 0) = 4 (9) - 0 f (9, 0) = 36 f (0, 7) = 4 (0) - 7 f (0, 7) = - 7 minimum$

This problem has one minimum at (0, 7). This system is unbounded, so there is no maximum.

Computer Concepts plans to build new computer chips with different materials, Type A and Type B. The Type A material weighs x grams. The Type B material weighs y grams. The maximum weight for the computer chips is 250 grams. Type A costs $150 per gram, while Type B costs $200 per gram. Local investors will invest no more than $45,000 in materials.

1. 1. Write and graph the system of inequalities to find the vertices.

Type A yields 600 mb of storage per gram, and Type B yields 750 mb of storage per gram.

Determine the optimization equation for the amount of Type A and Type B material that Computer Concepts should build to maximize storage capacity.
Find the optimized number of Type A and Type B material to provide the highest storage capacity for Computer Concepts.

$(x, y) : (Type A, Type B) x + y \leq 250 150 x + 200 y \leq 45, 000 x \geq 0 y \geq 0 f (x, y) = 600 x + 750 y$
$f (0, 0) = 600 (0) + 750 (0) f (0, 0) = 0 f (100, 150) = 600 (100) + 750 (150) f (100, 150) = 172, 500 maximum f (0, 225) = 600 (0) + 750 (225) f (0, 225) = 168, 750 f (250, 0) = 600 (250) + 75 (0) f (250, 0) = 150, 000$
Computer Concepts should use 100 grams of Type A material and 150 grams of Type B material to have a maximum storage capacity of 172,500 mb.

Note

The final sentence in the problem provides the objective. Remember that you need to write this as “f (x, y) =”

Friendship INC is creating a new set of rings for a special event. It takes x grams of silver and y grams of gold to make a ring. All rings have a combination of gold and silver. The “traditional” ring has twice as much gold as silver and weighs no more than 4 grams. The “contemporary” ring has twice as much silver as gold and weighs no more than 5 grams. The rings need a minimum of 0.5 grams of silver and a minimum of 0.5 grams of gold. Write and graph the system of inequalities to find the vertices.

The profit earned per gram of silver is $6 per gram and $8 per gram of gold.

How much of each type of metal should they use in their rings to optimize profit?

$f (x, y) = 6 x + 8 y f (0.5, 0.5) = 6 (0.5) + 8 (0.5) f (0.5, 0.5) = 7 minimum f (0.5, 1.75) = 6 (0.5) + 8 (1.75) f (0.5, 1.75) = 17 f (2, 1) = 6 (2) + 8 (1) f (2, 1) = 20 maximum f (2.25, 0.5) = 6 (2.25) + 8 (0.5) f (40, 0) = 17.5$

$(x, y) : (silver, gold) 2 x + y \leq 5 x + 2 y \leq 4 x \geq 0.5 y \geq 0.5 f (x, y) = 6 x + 8 y$

Friendship INC should use 2 grams of silver and 1 gram of gold to maximize profits.