Inequalites 2

amathproblemthatneedsolve

amathproblemthatneedsolve

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189
Question:
Symon makes two types of cake; chocolate and carrot. he is able to bake 50 cakes at the most each week. • It takes him30 minutes to prepare each chocolate cake and 35 minutes to prepare each carrot cake. • Symon has 1620 minutes [27 hours] available to prepare these cakes per week. • he has a regular order for 12 chocolate and 10 carrot cakes each week that she must deliver. Symon makes a profit of $12 from each chocolate cake and $16 dollars from each carrot cake.

• the maximum profit that Symon can make.
• the number of chocolate cakes and carrot cakes that Symon needs to sell to maximize the profit.

My working:
Variables let x by chocolate cakes and y be carrot

Constraints, x+y≤50 (he is able to bake 50 cakes at most each week)
30x+35y≤1620 (the time it takes to make one chocloate cake + carrot cake needs to be less than or equal too 1620minutes)
x≥ 12 (he needs to make a regular order of 12 choc each week)
y≥ 10 (carrot needs to be 10 each week)

Profit=12x+16y

from that my points are
PointsProfit
A (12,38)$752
B (26,24$696
C (12,36)$720

from this I guess point A is the best with max profit of $752 and you sell 12 chocolate cakes and 36 carrot cakes. Have I done this correct?
 
I did not get those corner points (I got 2 out of 4 of them).
Can you show us your graph along with your work so we can find your mistakes?
 
Last edited:
amathproblemthatneedsolve said:
(26,24)? Is there a better site to graph this?
Click to expand...
Can't you see that the graph you have does NOT show the feasible region? As my friend Subhotosh Khan suggested, use paper and draw the lines. Regardless of what you might have been told you should shade away the bad points for each inequality. This way in the end the feasible region will be the blank space. Otherwise you will have to find where the four inequalities overlap!
 
Jomo said:
Can't you see that the graph you have does NOT show the feasible region? As my friend Subhotosh Khan suggested, use paper and draw the lines. Regardless of what you might have been told you should shade away the bad points for each inequality. This way in the end the feasible region will be the blank space. Otherwise you will have to find where the four inequalities overlap!
Click to expand...
I've done this now. my points(vertexes) are (12,36), (40,10), (12,10) also x+y=50 and 30x+35y=1620 cross at (26,24) but the most profit comes from (12,36) $720
 
I do not particularly like these questions. They are "fake" examples of an important real-world application. Real world problems are almost never soluble by graphing and frequently require integer programming rather than linear programming: you cannot bake 1/17th of a cake. Moreover, solving by finding vertices is, as jomo pointed out, frequently burdensome. Nevertheless, it is always possible in principle to solve through finding feasible vertices.

Your set-up was fine.

[MATH]x = \text {# of chocolate cakes.} [/MATH]
[MATH]y = \text {# of carrot cakes.} [/MATH]
[MATH]p = \text {profit.}[/MATH]
[MATH]p = 12x + 16y.[/MATH]
Constraints

[MATH]12 \le x,\ 10 \le y,\ x + y \le 50,\ \text { and } 30x + 35x \le 1620.[/MATH]
With four lines, you have at most 6 vertices. Actually, here, there are only 5. How many are "feasible"?

[MATH]x = 12 \ \& \ y = 10 \implies 12 + 10 < 50 \ \& \ 30 * 12 + 35 * 10 = 710 < 1620.[/MATH]
Feasible.

[MATH]x = 12 \ \& \ 30x + 35y = 1620 \implies y = \dfrac{1620 - 360}{35} = 36 > 10 \ \& 12 + 36 < 50.[/MATH]
Feasible.

[MATH]y = 10 \ \& \ x + y = 50 \implies x = 40 > 12 \ \& \ 30 * 40 + 35 * 10 = 1550 < 1620.[/MATH]
Feasible.

[MATH]y = 10 \ \& \ 30x + 35y = 1620 \implies x = \dfrac{1620 - 350}{30} \approx 42.3 > 10 \text { but } x + 10 > 50.[/MATH]
Not feasible.

[MATH]x = 12 \ \& \ x + y = 50 \implies y = 38 > 12 \text { but } 30 * 12 + 35 * 38 = 1690 > 1620.[/MATH]
Not feasible (for two reasons).

As Jomo said, in this case, graphing is a lot simpler, but it is not generally applicable if there are more than two variables. But, however, you did it, I too find three feasible vertices at (12, 10), (12, 36), and (40, 10).

Now you just compute profits at all 3. But it is obvious by inspection that

[MATH]12 * 12 + 10 * 16 < 12 * 12 + 36 * 16 = 720.[/MATH]
[MATH]40 * 12 + 10 * 16 = 640.[/MATH]
Well done. It is interesting that the maximum profit does not come from baking the maximum number of cakes, which is a bit counter-intuitive.
 
JeffM said:
I do not particularly like these questions. They are "fake" examples of an important real-world application. Real world problems are almost never soluble by graphing and frequently require integer programming rather than linear programming: you cannot bake 1/17th of a cake. Moreover, solving by finding vertices is, as jomo pointed out, frequently burdensome. Nevertheless, it is always possible in principle to solve through finding feasible vertices.

Your set-up was fine.

[MATH]x = \text {# of chocolate cakes.} [/MATH]
[MATH]y = \text {# of carrot cakes.} [/MATH]
[MATH]p = \text {profit.}[/MATH]
[MATH]p = 12x + 16y.[/MATH]
Constraints

[MATH]12 \le x,\ 10 \le y,\ x + y \le 50,\ \text { and } 30x + 35x \le 1620.[/MATH]
With four lines, you have at most 6 vertices. Actually, here, there are only 5. How many are "feasible"?

[MATH]x = 12 \ \& \ y = 10 \implies 12 + 10 < 50 \ \& \ 30 * 12 + 35 * 10 = 710 < 1620.[/MATH]
Feasible.

[MATH]x = 12 \ \& \ 30x + 35y = 1620 \implies y = \dfrac{1620 - 360}{35} = 36 > 10 \ \& 12 + 36 < 50.[/MATH]
Feasible.

[MATH]y = 10 \ \& \ x + y = 50 \implies x = 40 > 12 \ \& \ 30 * 40 + 35 * 10 = 1550 < 1620.[/MATH]
Feasible.

[MATH]y = 10 \ \& \ 30x + 35y = 1620 \implies x = \dfrac{1620 - 350}{30} \approx 42.3 > 10 \text { but } x + 10 > 50.[/MATH]
Not feasible.

[MATH]x = 12 \ \& \ x + y = 50 \implies y = 38 > 12 \text { but } 30 * 12 + 35 * 38 = 1690 > 1620.[/MATH]
Not feasible (for two reasons).

As Jomo said, in this case, graphing is a lot simpler, but it is not generally applicable if there are more than two variables. But, however, you did it, I too find three feasible vertices at (12, 10), (12, 36), and (40, 10).

Now you just compute profits at all 3. But it is obvious by inspection that

[MATH]12 * 12 + 10 * 16 < 12 * 12 + 36 * 16 = 720.[/MATH]
[MATH]40 * 12 + 10 * 16 = 640.[/MATH]
Well done. It is interesting that the maximum profit does not come from baking the maximum number of cakes, which is a bit counter-intuitive.
Click to expand...
I know it doesn't produce the maximum value but isn't point (26,24) also feasible?
 
What part are you not understanding? There are four corner points not three. Because you can find the max profit you will need that 4th point. Please find it.
JeffM said:
I do not particularly like these questions. They are "fake" examples of an important real-world application. Real world problems are almost never soluble by graphing and frequently require integer programming rather than linear programming: you cannot bake 1/17th of a cake. Moreover, solving by finding vertices is, as jomo pointed out, frequently burdensome. Nevertheless, it is always possible in principle to solve through finding feasible vertices.

Your set-up was fine.

[MATH]x = \text {# of chocolate cakes.} [/MATH]
[MATH]y = \text {# of carrot cakes.} [/MATH]
[MATH]p = \text {profit.}[/MATH]
[MATH]p = 12x + 16y.[/MATH]
Constraints

[MATH]12 \le x,\ 10 \le y,\ x + y \le 50,\ \text { and } 30x + 35x \le 1620.[/MATH]
With four lines, you have at most 6 vertices. Actually, here, there are only 5. How many are "feasible"?

[MATH]x = 12 \ \& \ y = 10 \implies 12 + 10 < 50 \ \& \ 30 * 12 + 35 * 10 = 710 < 1620.[/MATH]
Feasible.

[MATH]x = 12 \ \& \ 30x + 35y = 1620 \implies y = \dfrac{1620 - 360}{35} = 36 > 10 \ \& 12 + 36 < 50.[/MATH]
Feasible.

[MATH]y = 10 \ \& \ x + y = 50 \implies x = 40 > 12 \ \& \ 30 * 40 + 35 * 10 = 1550 < 1620.[/MATH]
Feasible.

[MATH]y = 10 \ \& \ 30x + 35y = 1620 \implies x = \dfrac{1620 - 350}{30} \approx 42.3 > 10 \text { but } x + 10 > 50.[/MATH]
Not feasible.

[MATH]x = 12 \ \& \ x + y = 50 \implies y = 38 > 12 \text { but } 30 * 12 + 35 * 38 = 1690 > 1620.[/MATH]
Not feasible (for two reasons).

As Jomo said, in this case, graphing is a lot simpler, but it is not generally applicable if there are more than two variables. But, however, you did it, I too find three feasible vertices at (12, 10), (12, 36), and (40, 10).

Now you just compute profits at all 3. But it is obvious by inspection that

[MATH]12 * 12 + 10 * 16 < 12 * 12 + 36 * 16 = 720.[/MATH]
[MATH]40 * 12 + 10 * 16 = 640.[/MATH]
Well done. It is interesting that the maximum profit does not come from baking the maximum number of cakes, which is a bit counter-intuitive.
Click to expand...
Hmm, I am finding four feasible vertices.
 
Jomo said:
What part are you not understanding? There are four corner points not three. Because you can find the max profit you will need that 4th point. Please find it.

Hmm, I am finding four feasible vertices.
Click to expand...
I thought I said 4 points..... (12,36), (40,10), (12,10) and (26,24)
 
JeffM said:
I do not particularly like these questions. They are "fake" examples of an important … application …
Click to expand...
Well, instructors need to start somewhere, and I remind you of something that you already know: 'important application' is not always the motivation, when assigning exercises.

JeffM said:
… I remember: a cow flew by.
Click to expand...
If that was a spherical cow, then your distraction is excused.

?
 
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