Hello.

Stephen struggling with maths from Kenya.

Happy to join.

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- Dec 7th 2013, 12:45 PMscwmbaHello
Hello.

Stephen struggling with maths from Kenya.

Happy to join. - Dec 7th 2013, 12:53 PMromsekRe: Hello
Hi Scwmba

- Dec 7th 2013, 01:00 PMscwmbaRe: 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.) - Dec 7th 2013, 03:44 PMromsekRe: Hello

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. - Dec 8th 2013, 08:20 AMscwmbaRe: Hello
Hello romsek.

(Greetings to you from a new member) - Dec 8th 2013, 08:29 AMromsekRe: Hello
Hi Scwmba, I hope you are having a pleasant day. Did you make any progress on yesterday's problem?

- Dec 8th 2013, 08:39 AMscwmbaRe: 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 - Dec 8th 2013, 08:42 AMscwmbaRe: Hello
Ive to leave but will be back tomorrow (kenyan time).

You're feedback & help is appreciated! - Dec 8th 2013, 08:46 AMromsekRe: Hello
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? - Dec 10th 2013, 06:13 AMscwmbaRe: Hello
Hello Romsek.

I'm back - and very glad indeed to see your response.

Right now, I'm reading through to understand... - Dec 10th 2013, 07:31 AMscwmbaRe: 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? - Dec 12th 2013, 11:24 PMromsekRe: Hello
- Dec 15th 2013, 12:16 AMscwmbaRe: Hello
Ive attached the plot from wolframalpha.com.

- Dec 15th 2013, 01:55 AMscwmbaRe: 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 - Dec 15th 2013, 02:07 AMromsekRe: 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

$\displaystyle P = 0.5 x + 0.4 y$

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

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

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.