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Sunny_sXe [5.5K]
3 years ago
5

Gymnast Clothing manufactures expensive soccer cleats for sale to college bookstores in runs of up to 500. Its cost (in dollars)

for a run of x pairs of cleats is: C(x) = 2950 + 6x + 0.1x2 (0 ≤ x ≤ 500). Gymnast Clothing sells the cleats at $130 per pair. Find the revenue and profit functions. How many should Gymnast Clothing manufacture to make a profit?
Mathematics
2 answers:
shtirl [24]3 years ago
6 0

Answer:

Manufacture more than 24 items to make a profit.

Step-by-step explanation:

Let's find the Revenue function:

Revenue (R) = Price per pair * Number of pairs sold

           R(x)=130x

Now let's find the profit function:

Profit (P(x)) = Revenue (R(x)) - Cost C(x)

(P(x))  = (R(x))  – C(x)

P(x)=130x-(2950+6x+0.1x^{2} ), Where, (0\leq x\leq 500)

The requirement to make a profit:

P(x)\geq 0

⇒130x-(2950+6x+0.1x^{2} )\geq 0

x=24.265    or    x=1215.735

Therefore, they should manufacture more than 24 items to make a profit.

Natalija [7]3 years ago
3 0

Revenue, R = 130x

Profit, P = Revenue – Cost

 P = R – C

P = 130x – 2950 + 6x + 0.1x^2

At P = 0

0 = -2950 + 136x +0.1x^2

X = 21.355, -1381.355

Therefore, manufacture greater than 21.355 items to make profit. 

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Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for reg
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Answer:

a) MAX--> PC (R,P) = 0,3R+ 0,5P

b) <u>Optimal solution</u>: 40.000 units of R and 10.000 of PC = $17.000

c) <u>Slack variables</u>: S3=1000, is the unattended demand of P, the others are 0, that means the restrictions are at the limit.

d) <u>Binding Constaints</u>:

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Step-by-step explanation:

I will solve it using the graphic method:

First, we have to define the variables:

R : Regular Gasoline

P: Premium Gasoline

We also call:

PC: Profit contributions

A: Grade A crude oil

• R--> PC: $0,3 --> 0,3 A

• P--> PC: $0,5 --> 0,6 A

So the ecuation to maximize is:

MAX--> PC (R,P) = 0,3R+ 0,5P

The restrictions would be:

1. 18.000 A availabe (R=0,3 A ; P 0,6 A)

2. 50.000 capacity

3. Demand of P: No more than 20.000

4. Both P and R 0 or more.

Translated to formulas:

Answer d)

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

To know the optimal solution it is better to graph all the restrictions, once you have the graphic, the theory says that the solution is on one of the vertices.

So we define the vertices: (you can see on the graphic, or calculate them with the intersection of the ecuations)

V:(R;P)

• V1: (0;0)

• V2: (0; 20.000)

• V3: (20.000;20.000)

• V4: (40.000; 10.000)

• V5:(50.000;0)

We check each one in the profit ecuation:

MAX--> PC (R,P) = 0,3R+ 0,5P

• V1: 0

• V2: 10.000

• V3: 16.000

• V4: 17.000

• V5: 15.000

As we can see, the optimal solution is  

V4: 40.000 units of regular and 10.000 of premium.

To have the slack variables you have to check in each restriction how much you have to add (or substract) to get to de exact (=) result.  

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