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Math Help - Hello

  1. #1
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    Hello

    Hello.
    Stephen struggling with maths from Kenya.
    Happy to join.
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  2. #2
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    Re: Hello

    Hi Scwmba
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  3. #3
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    Re: Hello

    There is a Linear Programming problem i need help in undetstanding how to solve (below).
    Which is best way in this forum please?


    A school store sells slide rules at a profit of 50 each and sweatshirts at a profit of 40 each.

    It takes two minutes of a salesgirl's time and two minutes of a cashier's time to sell a slide rule.

    It requires three minutes of a salesgirl's time but only one minute of a cashier's time to sell a sweatshirt.

    The school store operates for a maximum of two hours during a school day, and during this time there is one cashier and there are two salesgirls available for handling sales of the above items.

    How many of each item should the store attempt to sell each day in order to earn a maximum profit?

    (Answer: 30, 60; $39.)
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  4. #4
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    Re: Hello

    Quote Originally Posted by scwmba View Post
    There is a Linear Programming problem i need help in undetstanding how to solve (below).
    Which is best way in this forum please?


    A school store sells slide rules at a profit of 50 each and sweatshirts at a profit of 40 each.

    It takes two minutes of a salesgirl's time and two minutes of a cashier's time to sell a slide rule.

    It requires three minutes of a salesgirl's time but only one minute of a cashier's time to sell a sweatshirt.

    The school store operates for a maximum of two hours during a school day, and during this time there is one cashier and there are two salesgirls available for handling sales of the above items.

    How many of each item should the store attempt to sell each day in order to earn a maximum profit?

    (Answer: 30, 60; $39.)

    You should start by deriving your profit function, and 2 constraint equations from what you've written above.

    One constraint equation is based on cashier time, the other on salesgirl time.

    These 3 are all functions of n_rule, and n_shirt, i.e. how many of each you sell.

    see if you can get this far.
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  5. #5
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    Re: Hello

    Hello romsek.
    (Greetings to you from a new member)
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  6. #6
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    Re: Hello

    Hi Scwmba, I hope you are having a pleasant day. Did you make any progress on yesterday's problem?
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  7. #7
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    Re: Hello

    Thanks for you're tip.
    This is how far i've gone:

    Let
    x–number of slide rules sold
    y–number of sweatshirts sold
    P–profit finction.

    P= 0.50x + 0.40y

    subject to:

    Salesgirl time constraint-
    2 minutes to 1 sliderule. 120 minutes to 60 sliderules.
    3 minutes to 1 sweatshirt. 120 minutes to 40 sliderules.

    Cashier time constraint-
    2 minutes to 1 sliderule. 120 minutes to 60 sliderule.
    1 minute to 1 sweatshirt. 120 minutes to 120 sweatshirts.

    x,y=>0

    Im stuck
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  8. #8
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    Re: Hello

    Ive to leave but will be back tomorrow (kenyan time).
    You're feedback & help is appreciated!
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  9. #9
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    Re: Hello

    Quote Originally Posted by scwmba View Post
    Thanks for you're tip.
    This is how far i've gone:

    Let
    x–number of slide rules sold
    y–number of sweatshirts sold
    P–profit finction.

    P= 0.50x + 0.40y

    subject to:

    Salesgirl time constraint-
    2 minutes to 1 sliderule. 120 minutes to 60 sliderules.
    3 minutes to 1 sweatshirt. 120 minutes to 40 sliderules.

    Cashier time constraint-
    2 minutes to 1 sliderule. 120 minutes to 60 sliderule.
    1 minute to 1 sweatshirt. 120 minutes to 120 sweatshirts.

    x,y=>0

    Im stuck
    Ok. We need to convert your constraint equations into something more than just their definitions by using the fact that the store is only open for 2 hrs, and that there is 1 cashier and 2 salesgirls.

    In a sales day there are effectively 2 cashier hours, or 120 cashier minutes.

    There are 2*2 = 4 salesgirl hours or 240 salesgirl minutes.

    thus using your x and y above we get

    2 x + y < 120 is the cashier time constraint. Do you understand where all this came from?

    2x + 3y < 240 is the salesgirl time constraint. Again make sure you understand why.

    Now plot this region so you can see what it looks like. Do you have the tools to post a picture of your plot?
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  10. #10
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    Re: Hello

    Hello Romsek.
    I'm back - and very glad indeed to see your response.
    Right now, I'm reading through to understand...
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  11. #11
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    Re: Hello

    "The school store operates for a maximum of two hours during a school day, and during this time there is one cashier and there are two salesgirls available for handling sales of the above items."

    (time cashier uses to sell) + (time salesgirls use to sell) = (total time to sell)
    (cashier units sold per minute*units sold) + (salesgirls units sold per minute*units sold) = (total time to sell)

    No. Not right.

    Salesgirls time constraint:
    total time = (minutes to sell unit) * (units sold) <= (2*120)
    = [(minutes to sell unit x) * (units x sold)] + [(minutes to sell unit y) *(units y sold)] <= 240
    = [(2 min/x) * (x)] + [(3 min/y) * (y)] <= 240
    = 2x + 3y <= 240.

    Cashier time constraint:
    2x + 1y <= 120.

    p = 0.5x + 0.4y
    s.t.
    2x + 3y <= 240
    2x + y <= 120
    x,y => 0

    Thank you, Romsek !
    I'm getting the approach !
    This forum looks very promising for me.

    Which tool do you recommend for me to post picture of my plot?
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  12. #12
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    Re: Hello

    Quote Originally Posted by scwmba View Post
    "The school store operates for a maximum of two hours during a school day, and during this time there is one cashier and there are two salesgirls available for handling sales of the above items."

    (time cashier uses to sell) + (time salesgirls use to sell) = (total time to sell)
    (cashier units sold per minute*units sold) + (salesgirls units sold per minute*units sold) = (total time to sell)

    No. Not right.

    Salesgirls time constraint:
    total time = (minutes to sell unit) * (units sold) <= (2*120)
    = [(minutes to sell unit x) * (units x sold)] + [(minutes to sell unit y) *(units y sold)] <= 240
    = [(2 min/x) * (x)] + [(3 min/y) * (y)] <= 240
    = 2x + 3y <= 240.

    Cashier time constraint:
    2x + 1y <= 120.

    p = 0.5x + 0.4y
    s.t.
    2x + 3y <= 240
    2x + y <= 120
    x,y => 0

    Thank you, Romsek !
    I'm getting the approach !
    This forum looks very promising for me.

    Which tool do you recommend for me to post picture of my plot?
    Hi Schwmba, sorry for the delay in getting back to you. I didn't notice you had replied.

    goto wolframalpha.com and type in

    Plot[{y<=(240-2x)/3,y<=(120-2x)},{x,0,60}]

    exactly like that
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  13. #13
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    Re: Hello

    Ive attached the plot from wolframalpha.com.
    Attached Thumbnails Attached Thumbnails Hello-tmp_msp15451ggbc620ccc0d4580000694035hdg9111agd-1248939606.gif  
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  14. #14
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    Re: Hello

    x y s1 s2 p
    2 3 1 0 0 240
    2 1 0 1 0 120
    -0.5 -0.4 0 0 1 0


    x y s1 s2 p
    0 2 1 -1 0 120
    1 1/2 0 1/2 0 60
    0 -0.15 0 0.25 1 30


    x y s1 s2 p
    0 1 1/2 -1/2 0 60
    1 0 -1/4 1/4 0 30
    0 0 15/200 40/100 1 39


    s1=0
    s2=0
    x=30
    y=60
    p=39
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  15. #15
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    Re: Hello

    Hi Schwmba,

    Ok that plot shows you how many of each items you can sell in a day. The shop is physically able to sell x sliderules and y sweatshirts as long as x and y lie in the shaded area of your plot. So now you have to compute which x and y in that area maximizes your profit.

    If the gradient of your profit function equals 0 somewhere in that shaded area then there is a maximum or minimum at that point. So let's take the gradient of the profit function

    P = 0.5 x + 0.4 y

    \nabla{P}=\{\frac{\partial P}{\partial x},\frac{\partial P}{\partial y}\} (this is a vector with an x and y component)

    \nabla{P}=\{0.5,0.4\}

    and this doesn't equal 0 anywhere. So there is no maximum or minimum in the interior of our allowed area of x and y.

    Thus the maximum must lie on the boundary.

    We want to maximize profit so that means selling as many as possible so we will find the solution on the upper boundary of this area.

    So now for x and y along that top line, compute and plot the profit you make.
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