Optimization

Lesson 8.4 ended on a promise. The corners of a region are the points to check when one point must be picked as best. This lesson keeps that promise. Optimization asks not which values work but which working value is largest or smallest, and its answer is a single number. One variable comes first.

Problem

A number

x

satisfies

6 x + 11 \leq 95

. Solve for the bound the 8.3 way, then check the bound itself is allowed. Enter the largest value

x

can take.

Show a hint

Subtract $11$ from both sides, then divide by the positive $6$ . That gives a ceiling on $x$ , a number $x$ cannot exceed.
The ceiling is the answer only if $x$ can equal it. Put the ceiling back into $6 x + 11 \leq 95$ and see whether the inequality still holds.

Show the full solution

Subtracting

11

and dividing by

6

gives

x \leq 14

, and

x = 14

gives

6 (14) + 11 = 95

, which is allowed, so the largest value is

14

. The bound alone was not enough. The check that

14

is itself allowed is the second half of the answer, and the next problem shows why it can fail.

Problem

Two numbers

x

and

y

satisfy

x + y \leq 24

and

y \geq 7

. It is tempting to say

x

can reach

24

. Decide whether it actually can, then enter the largest value

x

can take.

Show a hint

Reaching $x = 24$ forces $y \leq 0$ . Check whether the constraint on $y$ allows that.
Rewrite the first constraint as $x \leq 24 - y$ , then make $y$ as small as the second constraint permits.

Show the full solution

Since

x \leq 24 - y

and

y

is at least

7

x \leq 17

, and

x = 17

y = 7

satisfies both constraints, so the largest value is

17

. The bound

24

fails because no allowed pair reaches it. A bound nothing reaches is not a maximum.

Problem

Markers cost 6 dollars each and Lena has 80 dollars to spend on them and nothing else. Enter the largest number of markers she can buy.

Show a hint

Write the constraint as $6 n \leq 80$ and solve for $n$ . The bound you get is not a whole number.
A count of markers must be a whole number. Step down to the nearest whole number and check that its cost fits inside the 80 dollars.

Show the full solution

From

6 n \leq 80

n \leq 13 \frac{1}{3}

, and since markers come in whole numbers the most is

13

, costing

78

dollars, so the answer is

13

. The bound

13 \frac{1}{3}

is not a possible count, since counts are whole numbers, and among whole numbers nothing larger than

13

stays within the

80

dollars.

Problem

A number

x

satisfies all three of

2 x \leq 30

x + 8 \leq 20

and

4 x \leq 50

. Each constraint alone caps

x

. Enter the largest value

x

can take.

Show a hint

Solve each of the three inequalities separately. You get three different ceilings on the same $x$ .
$x$ has to obey all three at once, so only the lowest of the three ceilings can be reached. Check that value against the other two constraints.

Show the full solution

The three constraints cap

x

15

12

, and

12 \frac{1}{2}

, and every allowed value must sit under all three, so the largest is the smallest cap,

12

. The cap

12 \frac{1}{2}

is close but looser. Only the tightest cap binds, and

12

passes the other two checks.

A region cut out by five boundary lines, one edge for each of

a

through

e

, with a dot at every corner. A linear expression changes steadily along any one edge, so no point in the middle of an edge beats both of its ends. Only the corners ever need checking, and on this region the best value sits at the gold corner.

With two variables the allowed values are no longer a stretch of the number line but one of 8.4's regions. Its corners sit where boundary lines cross, found by solving the pair as a system, chapter 5's work. A crossing counts as a corner only when it also satisfies every other constraint, since a point outside the region is not a corner of it.

Problem

One corner of a region is the point where the boundary lines

2 x + y = 25

and

x + 3 y = 30

cross. Solve the system and enter the

x

-coordinate of that corner.

Show a hint

A corner is just where two boundary lines meet, the system-solving from chapter 5. Solve the first equation for $y$ to set up a substitution.
Substituting $y = 25 - 2 x$ into $x + 3 y = 30$ leaves one equation in $x$ . Solve it, then check the pair in both boundary equations.

Show the full solution

Substituting

y = 25 - 2 x

into

x + 3 y = 30

gives

- 5 x = - 45

, so the corner is

(9, 7)

and the

x

-coordinate is

9

. Nothing new happened here. Finding a corner is exactly the line-crossing work of chapters 5 and 7.

Problem

The points satisfying

x \geq 0

y \geq 0

x + y \leq 10

and

3 x + y \leq 18

form a region. Enter how many corners this region has.

Show a hint

Cross the boundary lines in pairs, but keep only the crossings that satisfy every other constraint. Some crossings land outside the region.
Where $x + y = 10$ meets the $x$ -axis, check whether $3 x + y \leq 18$ still holds. A sketch the 8.4 way settles the count if it is unclear.

Show the full solution

The crossings that satisfy all four constraints are

(0, 0)

(6, 0)

(4, 6)

and

(0, 10)

, so the number of corners is

4

. The crossings

(10, 0)

and

(0, 18)

each fail the constraint they do not sit on. A crossing outside the region is not a corner of it.

The whole method fits in one pass. List the constraints, find every corner by crossing boundary pairs and keeping the crossings that satisfy everything else, evaluate the expression at each corner, and take the best. The winning value occurs at an actual point of the region, so both halves of the two-part answer are settled at once.

Problem

A region has corners

(0, 0)

(8, 0)

(6, 4)

and

(0, 7)

. Evaluate

3 x + 2 y

at all four corners and enter the largest value it takes on the region.

Show a hint

There is one evaluation per corner, four in all. Skipping any of them risks missing the largest value.
At $(8, 0)$ the $y$ term is zero. Compare that corner against $(6, 4)$ , where both terms are positive.

Show the full solution

The corner values are

0

24

26

and

14

, so the largest value of

3 x + 2 y

on the region is

26

. At

(8, 0)

the value is

24

, two less than at

(6, 4)

, which is why every corner needs to be checked.

Problem

The constraints

x \geq 2

y \geq 1

and

x + y \leq 9

cut out a triangle. Find its corners and enter the smallest value

3 x + 4 y

takes on the triangle.

Show a hint

The three corners come from crossing the boundaries in pairs, and two come cheap since two boundaries are the lines $x = 2$ and $y = 1$ .
A minimum sits at a corner for the same edge reason a maximum does. Evaluate $3 x + 4 y$ at all three corners and keep the smallest value.

Show the full solution

The corners are

(2, 1)

(8, 1)

and

(2, 7)

, where

3 x + 4 y

equals

10

28

and

34

, so the smallest value is

10

. Minimizing is the same table as maximizing. Only the end you read it from changes.

A corner can have fractional coordinates, but a count of objects must be whole. The value at such a corner is still a bound. Round the bound down, then find an allowed whole-number point that reaches it. Rounding the corner's coordinates instead can give a point outside the region or miss the best plan, and the next problem sets exactly that trap.

Problem

A snack stand makes

x

fruit cups and

y

lemonades, whole numbers of each. Supplies allow

x + y \leq 8

and prep time allows

3 x + y \leq 15

. Each fruit cup earns 2 dollars and each lemonade 1 dollar. Enter the largest number of dollars the stand can earn.

Show a hint

Find the corners of the region and evaluate the earnings at each one. The best corner has fractional coordinates, so its value is only a bound, not yet the answer.
Round that bound down to a whole number of dollars, then find whole numbers of cups and lemonades that satisfy both constraints and earn exactly that amount.

Show the full solution

The best corner is

(\frac{7}{2}, \frac{9}{2})

, where

2 x + y = 11 \frac{1}{2}

, so earnings are at most

11

, and

x = 3

y = 5

satisfies both constraints and earns

11

. Rounding the corner's coordinates down to

(3, 4)

earns only

10

and misses the best plan. Round the bound down, then look for an allowed point that reaches it.

Problem

Mara has 75 dollars to spend on cupcake boxes. Each box costs 8 dollars and holds 12 cupcakes, and her backpack has room for 60 cupcakes. Enter the largest number of boxes she can carry home.

Show a hint

Two constraints cap the number of boxes, one from the budget and one from the backpack. Work out both caps before trusting either.
Money allows $9$ boxes, but $9$ boxes hold $108$ cupcakes. Which cap is actually the smaller one?

Show the full solution

The budget allows

9

boxes but the backpack caps the count at

60 \div 12 = 5

, and the smaller cap is the real limit, so the most Mara can carry home is

5

. Money is the slack constraint here. The limit comes from whichever cap is smaller, not from whichever the problem states first.

The chapter comes together here. Each constraint is rearranged under 8.1's rules and solved into a bound the 8.3 way, the region and its corners are drawn the 8.4 way, and the best corner is picked out with this lesson's two-part answer. Two problems remain, and each uses all of it at once.

Problem

The constraints

x \geq 0

y \geq 0

3 x + y \leq 21

and

x + 2 y \leq 12

cut out a four-cornered region. Find all four corners, evaluate

5 x + 4 y

at each, and enter the largest value it takes.

Show a hint

Cross the boundary lines in pairs and discard any crossing that violates another constraint. Exactly four corners survive.
The corners on the axes come quickly. The last corner is where $3 x + y = 21$ meets $x + 2 y = 12$ , a chapter 5 system. Then evaluate $5 x + 4 y$ at all four.

Show the full solution

The corners are

(0, 0)

(7, 0)

(6, 3)

and

(0, 6)

, since the crossings

(12, 0)

and

(0, 21)

each fail the constraint they do not sit on, and

5 x + 4 y

takes the values

0

35

42

and

24

, so the largest value is

42

. Evaluating at all four corners is the whole check, since the largest value must occur at one of them.

Problem

Whole numbers

x \geq 0

and

y \geq 0

satisfy

2 x + y \leq 16

and

x + 3 y \leq 21

. Enter the largest value the expression

7 x + 6 y

can take over all such pairs.

Show a hint

Run the corner method to get the bound. The best corner is not a whole-number point, so the bound is not attained.
Fix $x$ at each whole value from $0$ to $8$ , take the largest $y$ both constraints allow, and evaluate $7 x + 6 y$ . Keep the best of the nine columns.

Show the full solution

The best corner is

(\frac{27}{5}, \frac{26}{5})

, where

7 x + 6 y = 69

, so

69

is a bound but no whole-number point reaches it, and scanning each column,

x

from

0

8

with the largest allowed

y

, finds the best allowed point

(6, 4)

, where

7 x + 6 y = 66

. The scan misses nothing, since for each

x

the expression is largest at the biggest allowed

y

, and the best of the nine columns is

66

Every boundary in this chapter was a straight line because every expression was linear. Chapter 9 turns to expressions where a letter is multiplied by itself. Lesson 9.1 opens Quadratics I with factoring, writing $x^{2} + 7 x + 12$ as $(x + 3) (x + 4)$ , two linear factors, so the straight-line work just finished stays in use.

Practice these ideas

Practice

Of all the numbers satisfying

x + 6 \leq 25

, enter the largest.

Show the solution

Subtracting 6 gives

x \leq 19

, and

19 + 6 = 25

is allowed, so the largest is

19

Practice

A number

y

satisfies

5 y + 9 \leq 109

. Solve, check the bound is allowed, and enter the largest value of

y

Show the solution

5 y \leq 100

gives

y \leq 20

, and

y = 20

gives exactly

109

, so the largest value is

20

Practice

Erasers cost 4 dollars each and Ben has 35 dollars. Enter the largest number of erasers he can buy.

Show the solution

4 n \leq 35

gives

n \leq 8 \frac{3}{4}

, and erasers come whole, so Ben buys

8

. The check matters,

8

erasers cost

32

dollars and a ninth would break the budget.

Practice

Enter the largest value of

x

that satisfies

- 2 x \geq 10

Show the solution

Dividing by

- 2

flips the inequality to

x \leq - 5

, and

x = - 5

gives exactly

10

, so the largest is

- 5

. The flip is 8.1's rule, and forgetting it points the ray the wrong way.

Practice

Enter the smallest number satisfying

x - 7 \geq \frac{5}{2}

, as an exact fraction or decimal.

Show the solution

Adding 7 gives

x \geq \frac{19}{2}

, and

x = \frac{19}{2}

satisfies the original exactly, so the smallest value is

\frac{19}{2}

Practice

An integer

n

satisfies

4 n < 66

. Enter the largest possible value of

n

Show the solution

4 n < 66

gives

n < 16.5

, and the largest integer below the cutoff is

16

. The check

4 \cdot 16 = 64 < 66

works, and the answer would be the same with

\leq

since 16.5 is not an integer.

Practice

Raffle tickets cost 8 dollars each and a class must raise at least 210 dollars. Enter the smallest number of tickets that reaches the goal.

Show the solution

8 n \geq 210

gives

n \geq 26 \frac{1}{4}

, and tickets come in whole numbers, so the smallest count that reaches the goal is

27

. The bound is rounded up because it is not whole. Selling 26 raises only 208 dollars and misses the goal.

Practice

Candles cost 6 dollars each, the budget is 150 dollars, and the gift box holds three rows of seven candles. Enter the largest number of candles that can be bought and boxed.

Show the solution

Money allows

25

candles but the box holds only

21

, and the smallest cap binds, so the answer is

21

. The reached half still checks out, since

21

candles cost

126

dollars, inside the budget.

Practice

The boundary lines of

x + y \leq 11

and

3 x - y \leq 10

cross at a corner of the region where both hold. Enter the

x

-coordinate of that corner, as an exact fraction or decimal.

Show the solution

Adding the boundary equations gives

4 x = 21

, so the corner has

x = \frac{21}{4}

. Substituting back into

x + y = 11

gives

y = \frac{23}{4}

, and a corner with fractional coordinates is nothing unusual.

Practice

How many corners does the region

x \geq 0

y \geq 0

x + y \leq 6

have?

Show the solution

The three crossings

(0, 0)

(6, 0)

and

(0, 6)

all lie in the region, so the count of corners is

3

Practice

A region has corners

(0, 0)

(7, 0)

(5, 6)

and

(0, 9)

. Enter the largest value of

2 x + 3 y

on the region.

Show the solution

The corner values are

0

14

28

and

27

, so the largest value is

28

. The close second is

(0, 9)

with

27

, so the corner with the biggest

y

coordinate is not the best one here.

Practice

A region has corners

(3, 7)

(6, 4)

and

(9, 9)

. Enter the smallest value of

3 x + 2 y

on the region.

Show the solution

The corner values are

23

26

and

45

, so the smallest value on the region is

23

Practice

A region has corners

(0, 0)

(8, 0)

(5, 7)

and

(0, 9)

. The expression

9 x + 4 y

is largest at one corner. Enter that corner's

y

-coordinate.

Show the solution

The corner values are

0

72

73

and

36

, so the largest is

73

, at

(5, 7)

, and that corner's

y

-coordinate is

7

. The question asks for the coordinate itself, not the value

73

Practice

The points satisfying

x \geq 0

y \geq 0

x \leq 5

y \leq 5

x + y \geq 2

and

x + y \leq 8

form a region. Enter how many corners it has.

Show the solution

The boundary passes through

(2, 0)

(5, 0)

(5, 3)

(3, 5)

(0, 5)

and

(0, 2)

, so the count of corners is

6

. Cutting off a corner of the square trades one vertex for two, and that happens at both

(0, 0)

and

(5, 5)

, so the square's four corners become six.

Practice

Find the corners of the region satisfying

x \geq 0

y \geq 0

x + y \leq 10

and

4 x + y \leq 25

, then enter the largest value

3 x + 4 y

takes on the region.

Show the solution

The corners are

(0, 0)

(\frac{25}{4}, 0)

(5, 5)

and

(0, 10)

, with values

0

\frac{75}{4}

35

and

40

, so the largest is

40

. The largest value occurs at an axis corner, not at the crossing of the two slanted lines.

Practice

Find the corners of the region satisfying

x \geq 3

y \geq 2

and

x + y \leq 12

, then enter the smallest value of

6 x + 7 y

on that region.

Show the solution

The corners are

(3, 2)

(10, 2)

and

(3, 9)

, with values

32

74

and

81

, so the smallest value is

32

Practice

Whole numbers

x \geq 0

and

y \geq 0

satisfy

3 x + 2 y \leq 19

and

x + y \leq 7

. Enter the largest value

4 x + y

can take.

Show the solution

The best corner

(\frac{19}{3}, 0)

gives the bound

25 \frac{1}{3}

, no allowed whole-number pair gives a value of

25

, and at

(6, 0)

the value is

4 \cdot 6 + 0 = 24

. This is the step-down case, where the rounded bound of

25

has no reaching pair and the maximum is the next value an allowed pair reaches.