Quantitative Analysis for Business | Homework Helpers

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Need help with C1,C2,C3 and D See docs

QAT1 Task # 2

A.    X is the number of units of Brand X and Y is the number of units of Brand Y sold under the most favorable conditions

Each unit of Brand X contains (1) unit of nutrient, (2) units of color, (6) unit of flavor and (10) units of Grain.

Each unit of Brand Y contains (2) units of nutrient, (1) unit of color, (1) unit of flavor and (1) unit of grain.

Equation of profit

P = 100x +12.5y

Equation for nutrient

X*1+y*2

Equation for color

X*2+y*1

Equation for flavor

X*6+y*1

Equation for Grain

X*10+y*1

1.    With nutrient being in short supply no more than 500 units of nutrient can be used.  With a cap on nutrient this will be a maximum constraint.

X +2Y<=500

With color being in short supply no more than 150 units of color can be used.  With a cap on color this will be a maximum constraint.

2X+Y<=150

With flavor being in short supply no more than 400 units of flavor can be used.  With a cap on flavor this will be a maximum constraint.

6X+y<=400

With Grain being in short supply no more than 300 units of grain can be used.  With a cap on grain this will be a maximum constraint.

10X+y<=300

Therefore the four constraints are:

6x+y<=400

x+2y<=500

3x+y<=150

10x+y<=300

2.    The objective function is 100x +12.5y

B.    The profit line as shown in the graph passes through the point (0,200).  Therefore the total profit to be made can be obtained by using x = 0 and y = 200 in the profit function.

Total profit = 100*0 +12.5*200 = $2,500.

C.    Constraints are:

(1)    6x +y <=400

(2)    x+2y<=500

(3)    3x+y<=150

(4)    10x+y<=300

The feasible region for the problem is given by the region formed by the intersection of the graph of flavor, nutrient, grain and the x axis. We need to calculate the profits at the four points of intersection to find the optimal point.

We calculate the point of intersection for nutrient and grain.

(10X+y<+300) * 2 = (20X +2y = 600) €“ (x +2y = 500) = 19X = 100 or x = 100/19 = X = 5.26

Y = 300 €“ (10*5.26) = 247.4

X = 5.26

Y = 247.4

Profit using these values for x and y is: 100 * 5.26 +12.5 * 247.4 = $3,618.50

We calculate the point of intersection for flavors and nutrients

(6x  + y =400) * 2 = (12x + 2y = 800) €“ (1x+2y =500) = 11x = 300 or x = 27.27

Y = 400- (6 * 27.27) = 236.38

Y = 236.38

X = 27.27

Profit using these values for x and y is: 100 * 27.27 + 12.5 * 236.38 = $5,681.75

We calculate the point of intersection  for color and grains

(3x+y=150) * 2 = (6X +2y =300) €“ (10x + y =300)   =    -4x=0 = x = 4

Y = 150 €“ (3 * 4) = 138 = Y

Profit using values for x and y is:

100 * 4 + 12.5 * 138 = $2,125

The point of intersection for grains and x axis is: x = 300/10, y = 0

Total profit  = 100 * 30 +0 = $3,000.

Therefore maximum profit occurs at the intersection point of flavors and nutrients which is the optimal solution.

1. Optimal production of Brand X  is 27.27 units

2. Optimal production of Brand Y is 236.38 units

3. The feasible region was used to determine the possible solutions points.  Then putting the solution points in the profit function to determine the total profit at each point.   The optimal solution point was then determined which is the point that profit is optimal.

D.    Using the best production levels of Brand X and Y optimal profit would be $5,681.75