Start at the point where no drinks are made (so e = l = c = 0). Exchange one of these variables, in order to increase p. Repeat

Question

3 years ago

5

Start at the point where no drinks are made (so e = l = c = 0). Exchange one of these variables, in order to increase p. Repeat

the process of exchanging a basis variable to increase p until there are no exchanges which will make p larger. How many of each drink should be made?

Mathematics

1 answer:

3241004551 [841] · Answer 1 · 2021-11-30T08:50:56+03:00

Answer:

Coffee ground = 6969.5

Sugar = 0.75

Milk = 0.9375

Step-by-step explanation:

the given data can be written as follows

7 kg of coffee ground

1 gallon milk

10cups of sugar espressos

8 gms of ground 0.0625 gallons milk

0.125 cups of sugar Lattes

15 gms of ground no milk 0.125cups of sugar

CafeCubanos

7.5gms of ground no milk

0.125 cups of sugar

Given that, e represents the espressos, l represents the Lattes, and c represents cafe's cubanos.

Let g represents the amount of grounds, m represents the amount of milk, and s represents the amount of sugar.

price at espressos is $2, and $4 for lattes and $5 for Cafecubanos.

Let p be the amount of money they take in.

Let x , y ,z be the amounts of coffee ground, milk and sugar to be used for manufacture of espressos, lattes and cafcubanos.

So, the matrix form can be written as,

\begin{bmatrix} 8 & 0 &0 \\ 15& 0.0625 &0.125 \\ 7.5 & 0 & 0.125 \end{bmatrix}.\begin{bmatrix} x\\ y\\ z\end{bmatrix}=\begin{bmatrix} e\\ l\\ c\end{bmatrix}

So, this can be written as, 8x+0(y)+0(z)=e

15x+0.0625y+0.125z=l

7.5x+0(y)+0.125z=c

Total left over coffee ground when e=l=c=1 is

(7kg) -(8+15+7.5)=7000-30.5=6969.5

Total left over sugar is, 10 -(0.125+0.125)=1-0.250=0.75

Total left over milk is 1-0.0625=0.9375

So, for one quantitiy that is e=l=c=1, the left over quatitiy will be

6969.5g+9.750s+0.935m

NOw, cost of each espressos is $2

Cost of each lattes is $4 and cost of each cafe cabons is $5

So, p=2e+4l+5c