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jasenka [17]
2 years ago
11

Nate’s client said she wanted the width w of every room in her house increased by 2 feet and the length 2w decreased by 5 feet.

The polynomial (w+2)(2w−5) or 2w^2 −w−10 gives the new area of any room in the house. The current width of the kitchen is 16 feet. What is the area of the new kitchen?

Mathematics
1 answer:
Kay [80]2 years ago
6 0

Answer:

486 ft²

Step-by-step explanation:

For this, you want to replace every w with 16.

This gives (16+2)(2×16-5) or 2×16²-16-10.

I'm going to show this with the first example because the numbers are a bit easier.

You want to begin by working out the brackets. 16+2=18, and 2×16=32, giving us 18(32-5).

32-5=27, giving 18×27 as the equation.

This equals to 486, so the area is 486 ft².

**This content involves polynomials and area, which you might want to revise. I'm always happy to help!

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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
Contact [7]

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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Y= 3x - 10
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