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Math Help - Very Basic Algebra & Graph help - Profit.

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    Very Basic Algebra & Graph help - Profit.

    Firstly i'm sorry to bother all of you that will find this question really basic but I haven't studied Maths since GCSE 7 years ago and I am now doing a degree in Engineering without A Level Maths so any help would be great...I need to work out the algebra and the main issue is how to plot it on a graph!? I'm guessing I need to use y=mx + c but i'm not sure how to do that on excel.

    "
    A builder has a plot of land available on which he can build either luxury of standard houses.
    He decides to build at least 5 luxury and 10 standard houses. Planning restrictions limit him
    to no more than 30 houses altogether. A luxury house requires 300m2
    of land and a standard
    house 150m2
    . The plot is 6000m2
    . Profit is $12; 000 per luxury house and $8000 per standard
    house. How many of each type should he build to maximise his profit."

    Thank you anyone that can help!
    Adam
    Last edited by mr fantastic; November 12th 2011 at 01:59 PM. Reason: Title.
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    Re: Very Basic Algebra & Graph help - Profit (Sorry i'm new)

    ok so here is what I have worked out so far...

    Luxury Houses = x
    Basic Houses = y

    (300x + 5y) + (150y + 10x) = 6000m2 i'm not too sure if this is correct

    I also came to the conclusion for graph purposes the answer might be y= -0.5x + 10 ?

    I also know the most profit is 280,000 with 10 luxury houses and 20 basic.
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  3. #3
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    Re: Very Basic Algebra & Graph help - Profit (Sorry i'm new)

    Quote Originally Posted by adste89 View Post
    Firstly i'm sorry to bother all of you that will find this question really basic but I haven't studied Maths since GCSE 7 years ago and I am now doing a degree in Engineering without A Level Maths so any help would be great...I need to work out the algebra and the main issue is how to plot it on a graph!? I'm guessing I need to use y=mx + c but i'm not sure how to do that on excel.

    "
    A builder has a plot of land available on which he can build either luxury of standard houses.
    He decides to build at least 5 luxury and 10 standard houses. Planning restrictions limit him
    to no more than 30 houses altogether. A luxury house requires 300m2
    of land and a standard
    house 150m2
    . The plot is 6000m2
    . Profit is $12; 000 per luxury house and $8000 per standard
    house. How many of each type should he build to maximise his profit."

    Thank you anyone that can help!
    Adam
    1. Let x denote the number of standard houses;
    lety denote the number of luxury houses.

    2. Then you know:

    • x\geq 0~\wedge~y\geq 0
    • x+y\leq 30
    • 150x + 300y \leq 6000

    and the profit P is:
    P=8000x + 12000y

    3. Rearrange the inequalities and the equation into the form y \leq ... or y = ... and graph the first 4 inequalities. You'll get an irregular qudrilateral which represents the feasible area. (see attachment)

    4. Use the equation of P to get the maximum profit. I've drawn this case in red.
    Why exactly this line represents the maximum profit? What happens to the profit if you choose any other point of the feasible region? How can you decide that you've found the maximum case?
    Attached Thumbnails Attached Thumbnails Very Basic Algebra & Graph help - Profit.-maxhouse.png  
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    Re: Very Basic Algebra & Graph help - Profit (Sorry i'm new)

    Thanks for doing this...I can see the answer is correct I just don't understand how the graphs work so I can replicate it myself...why did you do x=-2/3x+70/3?
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  5. #5
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    Re: Very Basic Algebra & Graph help - Profit (Sorry i'm new)

    Quote Originally Posted by adste89 View Post
    Thanks for doing this...I can see the answer is correct I just don't understand how the graphs work so I can replicate it myself...why did you do y=-2/3x+70/3?
    From the equation

    P=8000x + 12000y

    you'll get: y = -\frac23 x + \frac P{12000}

    You now have to choose a point of the feasible region such that the y-intercept of this straight line is at it's maximum. That occurs at (20, 10). Plug in these values for x and y and solve for P. You'll get P = 280.000. And \frac{280.000}{12.000}=\frac{70}3. That's all.
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