A company manufactures two kinds of ice-cream. The vanilla ice-cream sells for $2.50 each while the chocolate flavour ice-cream sells for $4.50 cents each. It costs the company 1 labour hour to make the vanilla flavour ice-cream and 2 labour hours to make the chocolate flavour ice-cream. The company has a total of 300 labour hours available. It costs the company 3 machine hours for the vanilla ice-cream and 2 machine hours for the chocolate ice-cream. The company has a total of 480 machine hours available. How much of each type of ice-cream should the company produce to maximise revenue? What is the maximum revenue?
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Maximum and maximize revenue
- Thread starter Kenibu
- Start date
Hello,
I apologize in advance for my explanation, I hope you will find it helpful.
We will consider vanilla ice cream as x , and the choc ice cream as y. The labor hours are x+2y<=(smaller or equal)300
Machine hours are 3x+2x<=480
We are trying to find the maximum profit. First we find all the interception points from the labor hours if we consider x as 0, y= 150. and if y=0, x=300.
Therefore, we have 2 interception points (0,180) and (300,0). From machine hours if we consider x as 0, y= 240, and if y=0, x=160. Next we find the interception point of these problems, you can do it by solving them as simultaneous equations, and we find the last interception point of (90,105). Now we try all of these points to see which one has the largest profit.
P= 2.5x+4.5y.
P(0,150)= 675
P(160,0)= 400
P(90,105)= 697.5
P(0,240)= 1080
P(300,0)= 750
Therefore, the maximum profit would be achieved from selling 240 choc ice creams.
Once again I apologize for my explanation, English is not my first language.
I apologize in advance for my explanation, I hope you will find it helpful.
We will consider vanilla ice cream as x , and the choc ice cream as y. The labor hours are x+2y<=(smaller or equal)300
Machine hours are 3x+2x<=480
We are trying to find the maximum profit. First we find all the interception points from the labor hours if we consider x as 0, y= 150. and if y=0, x=300.
Therefore, we have 2 interception points (0,180) and (300,0). From machine hours if we consider x as 0, y= 240, and if y=0, x=160. Next we find the interception point of these problems, you can do it by solving them as simultaneous equations, and we find the last interception point of (90,105). Now we try all of these points to see which one has the largest profit.
P= 2.5x+4.5y.
P(0,150)= 675
P(160,0)= 400
P(90,105)= 697.5
P(0,240)= 1080
P(300,0)= 750
Therefore, the maximum profit would be achieved from selling 240 choc ice creams.
Once again I apologize for my explanation, English is not my first language.
Question 3: Assume that S(0) = 100, the current price of the stock, and
that the ex dividend price is:
S(1) =
80 with probability 1
8
90 with probability 2
8
100 with probability 3
8
110 with probability 2
8
The company will pay out a constant dividend D (independent of the future
stock price). Compute D for which the expected return on stock would be 18%.
that the ex dividend price is:
S(1) =
80 with probability 1
8
90 with probability 2
8
100 with probability 3
8
110 with probability 2
8
The company will pay out a constant dividend D (independent of the future
stock price). Compute D for which the expected return on stock would be 18%.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Isn't this exactly the same problem as posted by Ashna under the title "Question" last April 28?