# I need to know the variables (x,y), inequality statements and the optimization statement for the following,?Rainbow Paints had 1000 gallons of yellow paint and 1250 gallons of blue paint. These two...

Rainbow Paints had 1000 gallons of yellow paint and 1250 gallons of blue paint. These two colors will be mixed and sold as two different shades of green. One mix contains half yellow and half blue. The second mix will contain one-third yellow and two-thirds blue. Rainbow Paints has enough production space to make 2150 gallons. Marketing analysis has shown that the public will buy at least 250 gallons of the first mix and 150 gallons of the second mix. The profit on the first mix is $7.82 and $6.67 for the second mix. How much of each type should be made to have a maximum profit? What is the maximum profit Rainbow Paints can make?

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### 3 Answers

1/2 x + 1/3 y= 1000 -equation 1

1/2 x + 2/3y = 1250 - equation 2

x+y= 2150 - equation 3

Profit= 7.82x +6.67Y - 4

(1) - (2)

- 1/3y= -250

y= 750

sub. y=750 into (2)

1/2 x + 2/3 (750) =1250

1/2 x + 500= 1250

1/2 x= 750

x= 1500

Maximum profit = 7.82*1500 + 650*6.67

= $16065.50

Resources:Yellow paint 1000gal.Red paint 1250 gal

Mixture1: Ratio of yellow:blue = (1/2):((1/2)

Mixture2:Ratio of yellow :Blue= (1/3):(2/3)

Let the Mixture1 and Mixture2 be x gal and y gallons.

The yellow paint in the two mixtures = x/2+y/3 gallons canot exceed the company's stock of 1000gal.Similarly the blue pain quantity in two mixtures x/2+2y/3 gal cannot exceed 1250 of stock. I.e.'

x/2+y/3<=1000 (1)

x/2+2y/3<=1250 (2)

By capcity of production, x+y =<2150 (3)

By demand first mix x>=250 (4)

and second mix y>=150. (5)

The profit P for the quantity of x and y at the rate of $7.82 per unit of x and $6.67 per unit of y is given by:

P= 7.82x+6.67y (6)

The object is to maximise the profit P and find the quantities x and y satisfying the conditions or constraints (1) to (5).

Solution:

From (1) and (2): Treating like ordinary equations we find x and y

(2)-(1): y/3=1250-1000==> y=750. So, **y<=750 is a possibility.**

From this treating like simultaneous equations, we get by substitution in(1): x/2=100-250=750=>x=1500

So,** x<=1500 is a possible solution**

From these two possible solutions we get

X+y<=1500+750=2250 or

**x+y<=2250** (7)

From (3)** x+y < =2150. **(8).

The conditions (7) and (8) should hold good and the satisfying solution should maximise profit,P.Therefore, we decide to keep x at **1500** without reducing the quantity as the rate of profit per unitof x is** $7.82 ** which is higher than that of y at **$6.66** and limiting the y units to **650 **as x+y=2150 at this value satisfying all the consraints, giving **best or maximum profit.**

P=7.82***1500**+ **650***6.67 = $16,065.50, he maximum profit.

Did this help?

Wow, this is a complex linear programing problem.

OK, you can always determine your variables by looking at what is being maximized/minimized. In this case I said that the 1st green mix is x and the second green mix is y. Since the market research says that at least 250 gallons of the first mix will be bought we can say x >= 250, and likewise with the second mix (y >= 150).

You also have two more restraints with the number of gallons of each paint. There's the equation for the yellow paint; (1/2)x + (1/3)y <= 1,000 and the equation for the blue paint; (1/2)x + (2/3)y <= 1250.

The profit equation that you want to maximize is going to be 7.82x + 6.67y = C.

Not sure if you need the equations sloved for y, so just let me know if you still need assistance with it. Good luck!