Feasible region is the unshaded area and satisfies: x + y 1000 2x + y 1500 3x + 2y 2400 x 0 and y 0 SOLUTION.

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x

y

Feasible region is the unshaded area and satisfies:

x + y 1000

2x + y 1500

3x + 2y 2400

x 0 and y 0

SOLUTION

- 200 200 400 600 800 1000 1200

- 200

200

400

600

800

1000

1200

1400

1600

x

y

Draw a straight line x + 0.8y = k.

O

A

B

C

D

Maximum income = £880 at (400, 600)

SOLUTION

Move a ruler parallel to this line until it reaches the edge of the feasible region.The furthest point you can move it to is point B.At B (400, 600) the value of the objective function is 880.

Mini whiteboards

If x = no. of litres of orange juiceand y = no. of litres of lemon juice

Express the need for lemon juice to be at least 30% of a mixed juice drink as a constraint inequality for linear programming.

0.30(x + y) ≤ y => 0.3x ≤ 0.7y => y ≥ 3/7 x

Mini whiteboards

If x = no. of litres of orange juiceand y = no. of litres of lemon juice

Express the need for orange juice to be at least 60% of a mixed juice drink as a constraint inequality for linear programming.

0.60(x + y) ≤ x => 0.6y ≤ 0.4x => x ≥ 1½ y

Mini whiteboards

If x = no. of tennis balls y = no. of footballs z = no. of basketballs

Express the need for basketballs to be at least 30% of the total number of balls as a constraint inequality for linear programming.

0.30(x + y + z) ≤ z => 0.3(x+y) ≤ 0.7z => z ≥ 3/7 (x+y)

Mini whiteboards

If x = no. of tennis balls y = no. of footballs z = no. of basketballs

Express the need for footballs to be at least 25% of the total number of balls as a constraint inequality for linear programming.

0.25(x + y + z) ≤ y => 0.25(x+z) ≤ 0.75y => y ≥ 1/3 (x+z)

Mini whiteboards

If x = no. of tennis balls y = no. of footballs z = no. of basketballs

Express the need for tennis balls to be at least one third of the total number of balls as a constraint inequality for linear programming.

1/3(x + y + z) ≤ x => 1/3(y+z) ≤ 2/3 x => x ≥ 1/2 (y+z)

In order to ensure optimal health (and thus accurate test results), a lab technician needs to the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of protien. But the rabbits should be fed no more than five ounces of food a day. Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y, and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of protein per ounce, and costs £0.20 per ounce. Food Y contains 12 g of fat, 12 g of carbohydrates, and 1 g of protein per ounce, at a cost of £0.30 per ounce.What is the optimal blend?

1.

2.

3.

Since the exercise is asking for the number of ounces of each food required for the optimal daily blend, my variables will stand for the number of ounces of each:x: number of ounces of Food Xy: number of ounces of Food YSince I can't use negative amounts of either food, the first two constrains are the usual ones: x > 0 and y > 0. The other constraints come from the grams of fat, carbohydrates, and protein per ounce:fat:        8x + 12y > 24 carbs:  12x + 12y > 36 protein:  2x +   1y >   4 Also, the maximum weight of the food is five ounces, so: x + y < 5The optimisation equation will be the cost relation C = 0.2x + 0.3y, but this time I'll be finding the minimum value, not the maximum.After rearranging the inequalities, the system graphs as:

(Note: One of the lines above is irrelevant to the system. Can you tell which one?)When you test the corners at (0, 4), (0, 5), (3, 0), (5, 0), and (1, 2), you should get a minimum cost of sixty pence per daily serving, using three ounces of Food X only.

Independent Study

• Topic Assessment in teacher resources/moodle

• Exam questions with markschemes also in teacher resources/moodle.