Maximization

this one is killing me. I have tried every possibility, but cannot get the answer.

A candy company has 150kg of chocolate-covered nuts and 90kg of chocolate covered raisins to be sold as two different mixes. One mix will contain half nuts and half raisins and will sell for $7 per kg. The other mix will contain 3/4 nuts and 1/4 raisins and will sell for $9.50 per kg.

a. How many kilograms of each mix shoould the company prepare for the maximum revenue? Find the maximum revenue.
b. The company raises the price of the second mix to $11 per kg. Now how many kilograms of each mix should the company prepare for the maximum revenue? Find the maximum revenue.

I don't need the answer. Just the formula. Here is what I have tried:

x < 150
y< 90
.5 x + .5 y = 7
.75 x + .25 y = 9.5
not sure what z is.

Hello, meminusgarfield!

I'll get you started . . .

A candy company has 150 kg of chocolate-covered nuts and 90 kg of chocolate-covered raisins to be sold as two different mixes.
One mix will contain half nuts and half raisins and will sell for $7 per kg.
The other mix will contain 3/4 nuts and 1/4 raisins and will sell for $9.50 per kg.

a. How many kilograms of each mix shoould the company prepare for the maximum revenue?
. . Find the maximum revenue.

Click to expand...

In Mix A, there is an equal amount of nuts and raisins.
Let \(\displaystyle x\) = kg of nuts = kg of raisins.

In Mix B, there is 3 times as much nuts as raisins.
Let \(\displaystyle y\) = kg of raisins, then \(\displaystyle 3y\) = kg of nuts.


We have this information:

. . \(\displaystyle \begin{array}{c||c|c||c|} & \text{Nuts} & \text{Raisins} & \text{Price} \\ \hline \hline \text{mix A} & x & x & \$7.00 \\ \hline \text{mix B} & 3y & y & \$9.50 \\ \hline \hline \text{Total:} & 150 & 90 \end{array}\)


We have these inequalties:

. . \(\displaystyle \begin{Bmatrix} x \:\ge\: 0 \\ y \:\ge\:0 \\x + 3y \:\le \: 150 \\ x + y \:\le \:90 \end{Bmatrix}\) .[1]


There will be \(\displaystyle x + x \,=\,2x\) kg of Mix A at $7.00 per kg.
. . The revenue from Mix A is: .\(\displaystyle 7(2x) \:=\:14x\text{ dollars.}\)

There will be \(\displaystyle 3y + y \,=\,4y\) kg of Mix B at $9.50 per kg.
. . The revnue from Mix B is: .\(\displaystyle 9.5(4y) \:=\:38y\text{ dollars.}\)

Hence, the total revenue is: .\(\displaystyle R \;=\;14x + 38y\) .[2]


Graph the region determined by [1] and locate its vertices.
. . You should get: .\(\displaystyle (0,0),\;(90,0),\;(0,50),\;(60,30)\)

Test the vertices into [2] and see which vertex produces maximum revenue.


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I've gone this far . . . I might as well answer the question . . .


. . \(\displaystyle \begin{array}{ccccccc} (0,0): & R &=& 14(0) + 38(0) & =& 0 & \text{Minimum revenue ... LOL!} \\ (90,0): & R &=& 14(90) + 38(0) &=& 1260 \\ (0,50): & R &=& 14(0) + 38(50) &=& 1900 \\ (60,30): & R &=& 14(60) + 38(30) &=& 1980& \Leftarrow\:\text{ Maximum revenue} \end{array}\)