Question

HELP! A bakery sells pies and cakes. Each pie takes 20 minutes to prepare and 45 minutes to cook. Each cake takes 30 minutes to cook and a total of 40 minutes to mix and decorate. In a given day, the bakery has 16 hours of employee time for food preparation and 3 ovens available for 8 hours of each. The bakery makes a profit of $5.50 on each pie and $6.00 on each cake.

a) Let x = the number of pies and y = the number of cakes the bakery makes in a given day. Translate a system of linear inequalities.
c) Find the number of each dessert the bakery should make each day to max its profit.

Answer 1

please, I'm trying....

Answer 2

I think, that we have three ovens for 16 hours, so it is like to have one oven which is availabile for 16*3=48 hours, do you agree?

Answer 3

because three ovens are available contemporarily

Answer 4

so 48 hours are equivalent to to 48*60 minutes, then your first inequality is:
\[65*x+70*y \le 48*60\]
do you agree?

Answer 5

@djibben615

Answer 6

so, after a simple simplification, I rewrite the first inequality as below:
\[13x+14y \le576\]

Answer 7

gotcha! What would the second system be?

Answer 8

@Michele_Laino

Answer 9

Now we have to write the second inequality do you agree?

Answer 10

I think that if our bakery sells only cakes, then he makes the maximum gain. Now we have to calculate maximum number of cake that our bakery can sell

Answer 11

I think that we can write:
\[70*y \le 48*60\]
so:
\[y \le \frac{ 48*60 }{ 70 }\le \frac{ 49*60 }{ 70 }=42\]
so our bakery can sell at maximum 42 cakes and can gain at maximum
42*6.00=252$

Answer 12

gotcha! this makes so much more sense now! thanks!

Answer 13

then our second inequality can be this:
\[5.5*x+6*y \le 252\]

Answer 14

gotcha that looks right now

Answer 15

\[5.5x + 6y <= 252\]

Answer 16

second part is slightly more subtle

Answer 17

What about the pies tho? Part c asks the number of both pies and cakes it'll make to max the profits?

Answer 18

because we have to maximize the subsequent quantity:
5.5x+6y

Answer 19

\[5.5x + 6y \le 252\] this is the second inequality of the system correct?

Answer 20

yes!

Answer 21

So these make up the system?: \[\left\{ 13x + 14 y \le 576 \right\} \left\{ 5.5 + 6y \le 252 \right\}\]

Answer 22

yes I think!

Answer 23

Okay, so now to work on C?

Answer 24

yes!

Answer 25

I'm not sure that system will work since the second inequality has both x and y in it too. We have to be able to substitute the second inequality into the first one.

Answer 26

I think that we have to use only the first inequality and graph it

Answer 27

oh really? Okay

Answer 28

please note that, we got maximum of both x and y, when we have:
\[13x+14y=576\]
or, in other words, when:
\[576-13x\]
is divisible by 14

Answer 29

I'm trying to find solution...

Answer 30

If we proceed by attempts, we will get:
x=12, y=30

Answer 31

How would i graph that then?

Answer 32

please you have to graph this function:
\[y=\frac{ 576-13x }{ 14 }\]
or:
\[y=\frac{ 288 }{ 7 }-\frac{ 13 }{ 14 }x\]

Answer 33

gotcha! Now, how to get max profits for both pies and cakes?

Answer 34

I think setting
x=12 and y=30

Answer 35

so..... how would i figure out the total? :) thanks

Answer 36

total is:
5.5*12+6.0*30=246.00$