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Mars2501 [29]
3 years ago
10

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

ular gasoline and $0.50 per gallon for premium gasoline. Each gallon of regular gasoline contains 0.3 gallons of grade A crude oil and each gallon of premium gasoline contains 0.6 gallons of grade A crude oil. For the next production period, Southern has 18,000 gallons of grade A crude oil available. The refinery used to produce the gasolines has a production capacity of 50,000 gallons for the next production period. Southern Oil's distributors have indicated that demand for the premium gasoline for the next production period will be at most 20,000 gallons. a. Formulate a linear programming model that can be used to determine the number of gallons of regular gasoline and the number of gallons of premium gasoline that should be produced in order to maximize total profit contribution. b. What is the optimal solution? c. What are the values and interpretations of the slack variables? d. What are the binding constraints?
Mathematics
1 answer:
Contact [7]3 years ago
3 0

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>:

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

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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In circle o,find the value of x, rounded to the nearest tenth.
Gemiola [76]

Answer:

\huge\boxed{x \approx 5.4}

Step-by-step explanation:

We can break down this problem by first realizing different parts of the circle.

  • The line which is 8 units long is a chord of the circle.
  • The line that is 3.6 is <em>almost</em> the radius of the circle
  • The line that x sits on is the radius.

With this, we can find out if we find the radius of the circle, we have our answer.

We should also note that the angle formed by the 3.6 units long line and the chord is a right angle.

<em>What we need is a way to find the radius of the circle</em><em>. This will get us x</em>. The radius of a circle will be the length of any line that starts from point O and ends at the circle edge.

If we draw a line connecting the end of the 3.6 line at point O to the end of the 8 unit long chord, we get a triangle! (Image attached for reference).

We can solve for the hypotenuse using the Pythagorean Theorem. This theorem states that:

  • \displaystyle a^2 + b^2 = c^2

Since we know one side is 3.6, we can use that as A. The second side will be 4 since the 3.6 line lies directly in the center of the chord = 8/2 = 4!

  • 3.6^2 + 4^2 = c^2
  • 12.96+16=c^2
  • 28.96=c^2
  • \sqrt{28.96} = c
  • 5.4 \approx c

Therefore, since this is the radius of the circle (also the hypotenuse), this can be said for any line that comes from point O onto the edge of the circle.

The line X does just that. Therefore, the value of x is also 5.4.

Hope this helped!

4 0
3 years ago
The given equation has been solved in the table.
Alina [70]

From the steps shown in the table given, we can conclude that: D. in solving the equation, the subtraction property of equality was not applied.

<h3>What is the Subtraction Property of Equality?</h3>

If a + b = c, to apply the subtraction property of equality so that b is moved to the other side of the equation, we would have:

a + b - b = c - b

a = c - b

Thus, from the table given, the none of the steps showed that the subtraction property of equality was applied in solving the equation, so, the answer is: D.

Learn more about the subtraction property of equality on:

brainly.com/question/1601404

#SPJ1

7 0
2 years ago
Somebody help me!
tiny-mole [99]

Answer:

josh vendió 17 libros recaudando 306 dolares .

jessica vendió  255 libros recaudando 4590 dolares.

Step-by-step explanation:

Sustituimos por variables :

libros que vendió Jessica = x

libros que vendió Josh = y

entonces:

x + y = 272

Jessica vendió 15 veces mas libros que josh:

x = 15y

Reemplazamos en la anteriior ecuacion:

15y +y = 272

16y = 272

y = 17

Reemplazamos en la primela ecuacion :

x + 17 = 272

x = 255

7 0
3 years ago
The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>

7 0
3 years ago
Rewrite the definition “Perpendicular lines form right angles” as a biconditional statement using “if and only if”
vovangra [49]
Well, to write the definition of "Perpendicular lines form right angel" using "if and only if". the sentence would become :
<em>
</em><em>Two lines are perpendicular if and only if they form a right angle
</em><em>
</em><em />Hope this helps


6 0
3 years ago
Read 2 more answers
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