help EXAM linear programming & objective functions

[emmylee]

please help me, i am in year 12 and my exam is in 2 days. i thought i had this topic downpat - but the revision sheet tells me i dont
here is a question (its quite lengthy but mainly just information)

soft toys are a sideline for a dressmaking factory using scraps from the main business - dresses.
thereis only enough waste from each dress to make 2 toys at most. dresses take 2 hours to make, while toys take 30 minutes each tomake.
there are 72 worker hours available each day. in order to satisfy regular customers, at least 20 dresses must be made each day.
find the MAXIMUM profit the factory can make if dresses return $30 and toys $8 each.

we have been taught at schoolto find the
VARIABLES: (D=dresses, T=toys)
PARAMETERS: dresses - $30 return, 2 hours labor & toys - $8 return, 30 min labor, 1D=2T.
CONSTRAINTS: below/equal to 72 hours (4320 minutes) and above/equal to 20 dresses

i now have to form an equation using all of this information. i came up with 120D + 30T below/equal to4320 minutes (for working time), however this is not including the information given that 1 dress=2 toys.
can you please help me so i can graph the equation :lol:
thankyou somuch for your time, your a godsend!!!

 

[wjm11]

soft toys are a sideline for a dressmaking factory using scraps from the main business - dresses.
thereis only enough waste from each dress to make 2 toys at most. dresses take 2 hours to make, while toys take 30 minutes each tomake.
there are 72 worker hours available each day. in order to satisfy regular customers, at least 20 dresses must be made each day.
find the MAXIMUM profit the factory can make if dresses return $30 and toys $8 each.

we have been taught at schoolto find the
VARIABLES: (D=dresses, T=toys)
PARAMETERS: dresses - $30 return, 2 hours labor & toys - $8 return, 30 min labor, 1D=2T.
CONSTRAINTS: below/equal to 72 hours (4320 minutes) and above/equal to 20 dresses

i now have to form an equation using all of this information. i came up with 120D + 30T below/equal to4320 minutes (for working time), however this is not including the information given that 1 dress=2 toys.

“120D + 30T below/equal to4320 minutes” is an incorrect assumption. It assumes that for every dress being made, two toys are being made. It is not necessary to make the toys, although that is a possibility. Let’s look at what we have:

D > or = 20

What if we made only dresses?

2D < or = 72, which gives us
D < or = 36

Now, if we only made the minimum number of dresses (20), that would take 40 hours, and we’d still have time to make 64 toys (maximum). The least number of toys we could make is zero.

T > or = 0
T < or = 64

If we graph this information with D on the Y axis and T on the x axis, it defines a rectangular area. We have one more piece of information however: we can make at most two toys for every dress, so

D > or = .5T

This line cuts off the lower right hand corner of our rectangular area. Thus, there are five critical points to be examined, the vertices of this five-sided figure. Plug those (T,D) values into the Profit eqn to find out which one maximizes profit:

P = 30D + 8T

I hope this helps.

 

[emmylee]

what about the equation?

i went back to have a look and tried to graph the contstraints you gave me although there is an extra component that we have learnt. we create an equation and then use the intercpt method (where D=0, where T=0), sub these values in and create our first line on the graph. this line is on an angle passing through both X and Y axis...
would this equation be:

120D + 30T < or = 4320 minutes (72 hrs)

thanks again

 

[wjm11]

120D + 30T < or = 4320 minutes (72 hrs)

You are correct emmylee. This will simplify to

D < or = (-.25)T + 36

which will further define our solution area and create one new critical point to test.

Sorry for the omission; I was getting sleepy last night when I posted.