A company produces two products, A and B. The sales volume for A is at least 80% of the total sales of both A and B. However, th
e company cannot sell more than 110 units of A per day. Both products use one raw material, of which the maximum daily availability is 300 lb. The usage rates of the raw material are 2 lb per unit of A, and 4 lb per unit of B. The profit units for A and B are $40 and $90, respectively. Determine the optimal product mix for the company.
1 answer:
Answer:
- 100 A
- 25 B
Explanation:
Per pound of raw material, B produces 90/4 = $22.50 in profit, while A produces $40/2 = $20 in profit. Clearly, B is more profitable, so should be produced in the maximum possible number. Of course, the limitation is ultimately the amount of available raw material.
The requirement that A be at least 80% of the product mix means that at least 4 A must be produced for each B. Then, for each B produced, we have a raw material utilization of ...
4A(2 lb/A) +1B(4 lb/B) = 12 lb
The maximum number of B that can be produced is 300 lb/(12 lb/B) = 25 B.
Production of 25 B and 100 A is the optimal product mix.
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100 A is below the maximum of 110 that can be sold, so that is not a limit in this scenario.
_____
The attachment shows a graphical solution to the inequalities ...
- x ≥ 0.80(x +y) ⇒ x ≥ 4y
- x ≤ 110
- 2x + 4y ≤ 300
- x ≥ 0, y ≥ 0
Quantities of A and B are represented by x and y, respectively.
The objective function 40x+90y is maximized when the line is at the vertex of the feasible region that puts it farthest from the origin. That point is (x, y) = (100, 25). A close candidate is the point (x, y) = (110, 20), for which the profit is slightly less.
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Answer: Heat transfer (Q) is 521 kJ.
Explanation: In the piston-cilinder assembly, we can suppose that the oxygen act as an ideal gas, so, it can be used the General Gas Equation:
PV=RT, where:
P is pressure;
V is volume;
m is mass;
M is molar mass;
R is a constant: R = 8.314. m³bar.K⁻¹.mol⁻¹;
T is temperature;
Using this equation, find the intial temperature:
PV = RT
1.2 = .8.314.
.T
T = 1.924. K
To determine the final temperature, use Combined Gas Law:
=
, in which, the left side of the equality is related to the initial values and the right side, to the final values.
As pressure is constant:
T =
T =
T = 3.85. K
With the temperatures, calculate the heat transfer of the process:
Q = m.k.ΔT, where:
k is heat constant
ΔT = T -
Q = m.k.ΔT
Q = 2.1.35.(3.85 - 1.92).
Q = 521 kJ
The heat transfer in the process is 521 kJ.
Answer:
See explanation for step by step procedure to get answer.
Explanation:
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The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.
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