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
for e=l=c=0, the price p is 0.
If e=l=c=1, then P=2+4+5=11
Similarly, by giving the values like this, we will get different value at different values of e,l,c
e=8x,
l=15x+0.0625y+0.125z
c=7.5x+0(y)+0.125z
Choose the values of x, y and in such a way that, x\leq7000,y\leq1,z\leq10
