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Stells [14]
3 years ago
6

Bullco blends silicon and nitrogen to produce two types of fertilizers. Fertilizer 1 must be at least 40% nitrogen and sells for

$70 per pound. Fertilizer 2 must be at least 70% silicon and sells for $40 per pound. Bullco can purchase up to 80 pounds of nitrogen at $15 per pound and up to 100 pounds of silicon at $10 per pound. Assuming all fertilizer products can be sold, formulate and solve an LP to help Bullco maximize profits.
Mathematics
1 answer:
leva [86]3 years ago
8 0

Answer:

Maximize

z = 70(Xs1 + Xn1 ) + 40(Xs2 + Xn2 ) - 10 (Xs1 + Xs2 ) - 15(Xn1 + Xn2 )  

Subject to the constraints  

Xs1 + Xs2 ≤ 100

Xn1 + Xn2 ≤ 80

Xn1 ≥ 0.4 ( Xs1 + Xn1 )

Xs2 ≥ 0.7 ( Xs2 + Xn2 )

All Variables ≥ 0

Step-by-step explanation:

Firstly lets consider Xs1 and Xs2 to be the number of pounds of silicon used in fertilizer1 and fertilizer2 respectively

Also let Xn1 and Xn2 be the number of pounds of nitrogen used in fertilizer1 and fertilizer2 respectively

We know that the objective is to maximize the profits of Bullco.

z = [(Selling price of fertilizer1) (Amount of silicon and nitrogen used to produce fertilizer1) + (Selling price of fertilizer2) (Amount of silicon and nitrogen used to produce fertilizer2) - (Cost of silicon) (Amount of silicon used to produce fertilizer I and 2) - (Cost of nitrogen) (Amount of nitrogen used to produce fertilizer I and 2)]

so

z = 70(Xs1 + Xn1 ) + 40(Xs2 + Xn2 ) - 10 (Xs1 + Xs2 ) - 15(Xn1 + Xn2 )  

Now

Constraint 1;  At most, 100 lb of silicon can be purchased

Amount of silicon used to produce fertilizer 1 and 2 ≤ 100

Xs1 + Xs2 ≤ 100

Constraint 2; At most, 80 lb of nitrogen can be purchased

Amount of nitrogen used to produce fertilizer 1 and 2 ≤ 80

Xn1 + Xn2 ≤ 80

Constraint 3; Fertilizer 1 must be at least 40% of nitrogen

Amount of nitrogen used to produce fertilizer 1 ≥ 40% (fertilizer 1)

Xn1 ≥ 0.4 ( Xs1 + Xn1 )

Constraint 4; Fertilizer 2 must be at least 70% of silicon

Amount of silicon used to produce fertilizer 2  ≥ 70% (fertilizer 2)

Xs2 ≥ 0.7 ( Xs2 + Xn2 )  

so the formulization of the given linear program is,  

Maximize

z = 70(Xs1 + Xn1 ) + 40(Xs2 + Xn2 ) - 10 (Xs1 + Xs2 ) - 15(Xn1 + Xn2 )  

Subject to the constraints  

Xs1 + Xs2 ≤ 100

Xn1 + Xn2 ≤ 80

Xn1 ≥ 0.4 ( Xs1 + Xn1 )

Xs2 ≥ 0.7 ( Xs2 + Xn2 )

All Variables ≥ 0

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Answer:

See below.

Step-by-step explanation:

PROBLEM A

The formula for the volume of a cylinder is

V = πr²h, where πr² is the area of the base.

V = area of base x height

  1. Choose any four numbers that are less than 10 for the height.
  2. Then, substitute into the formula and isolate "r". 10 = πr²h
  3. Double "r" to find the diameter.
  4. The area of the base is found (bolded) while you solve πr².

*I will use the calculator button for pi (π)

<u>Cylinder A: </u><u>h = 1</u>

10 = πr²h

10 = πr²(1)

10 = πr²                       Area of the base is 10m².

r = √ (10/π)

r = 1.78                        Radius is 1.78m.

d = 2r = 1.78*2 = 3.56                     Diameter is 3.56m.

<u>Cylinder B: </u><u>h = 2</u>

10 = πr²h

10 = πr²(2)

5 = πr²                       Area of the base is 5m².

r = √ (5/π)

r = 1.26                      Radius is 1.26m.

d = 2r = 1.26*2 = 2.52                     Diameter is 2.52m.

<u>Cylinder C: </u><u>h = 4</u>

10 = πr²h

10 = πr²(4)

2.5 = πr²                       Area of the base is 2.5m².

r = √ (2.5/π)

r = 0.89                        Radius is 0.89m.

d = 2r = 0.89*2 = 1.78                     Diameter is 1.78m.

<u>Cylinder D: </u><u>h = 5</u>

10 = πr²h

10 = πr²(5)

2 = πr²                       Area of the base is 2m².

r = √ (2/π)

r = 0.80                     Radius is 0.8m.

d = 2r = 0.80*2 = 1.6                     Diameter is 1.6m.

The answers are reasonable if the answer is close to 10 when you substitute the rounded numbers back into the formula and solve.

There are infinite possibilities because any numbers when used in the formula equates to 10 are possible. The height/radius can be any decimal number, which would make the other dimensions change.

PROBLEM B

Follow similar steps if you know the diameter of a cylinder. Use the same formula V = πr²h.

We need the radius, which is half the diameter. r = d/2 = 10/2 = 5

  1. Choose a random number for volume that divides easily by 25.
  2. Substitute "V" and "r²" (r² = 25).
  3. Isolate "h".
  4. The area of the base is 78.54 cm² every time (A = πr²).

<u>Cylinder A: </u><u>V = 100</u>

100 = π25h

4 = πh

h = 1.27                      The height is 1.27m.

<u>Cylinder B: </u><u>V = 200</u>

200 = π25h

8 = πh

h = 2.55                      The height is 2.55m.

<u>Cylinder C: </u><u>V = 75</u>

75 = π25h

3 = πh

h = 0.95                      The height is 0.95m.

<u>Cylinder D: </u><u>V = 125</u>

125 = π25h

5 = πh

h = 1.59                      The height is 1.59m.

The answers are reasonable if I can substitute two rounded values into the formula and get a number close to the third. The height is also a decimal number which occurs because of pi.

6 0
3 years ago
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