Inequalites

That is an excellent question. What have you done? Can you please post your work so that someone here can tell you how to continue with what you posted.
 
Let x = Children and y= Adults

x≥0
y≥0
x+y≤300
12x+3y≥900
4x+8y≥1000
y≥3x

P(profit)=10x+5y

so for maximum profit you use point (75,225) = $1875, 75 children and 225adults.

heres what I want to know if the adult and children expected profit changed to $7.50 (P=7.5x+7.5y) thats an decrease of 2.50 for children and an increase of 2.50 for adults but would be the new maximum profit? and how many child/adult would there be? do I use the same point 75,225 or another?
 
amathproblemthatneedsolve said:
Let x = Children and y= Adults

x≥0
y≥0
x+y≤300
12x+3y≥900
4x+8y≥1000
y≥3x

P(profit)=10x+5y

so for maximum profit you use point (75,225) = $1875, 75 children and 225adults.

heres what I want to know if the adult and children expected profit changed to $7.50 (P=7.5x+7.5y) thats an decrease of 2.50 for children and an increase of 2.50 for adults but would be the new maximum profit? and how many child/adult would there be? do I use the same point 75,225 or another?
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No, no. It is a brand new problem which has nothing to do with the other result of (75, 225). Do whatever you did to get (75,225) but with the new P function.
 
Jomo said:
No, no. It is a brand new problem which has nothing to do with the other result of (75, 225). Do whatever you did to get (75,225) but with the new P function.
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I got (75,225) by graphing

x+y≤300
12x+3y≥900
4x+8y≥1000
y≥3x

then putting the coordinate points within the feasible area into the P=10x+5y and the coordinates (75,225) gave me the maximum value. What do I need to do I'm confused?
 
OK, so maybe I was not correct saying this was a new problem. But certainly the max profit has nothing to do with the original P equation. Just like you found (75,225) to be the max for the original P (you found P for each corner point in the feasible set), do the same for the new P.
 
Jomo said:
OK, so maybe I was not correct saying this was a new problem. But certainly the max profit has nothing to do with the original P equation. Just like you found (75,225) to be the max for the original P (you found P for each corner point in the feasible set), do the same for the new P.
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That's the thing here it is https://www.desmos.com/calculator the best results I get by plugging in my points to P=7.5x+7.5y is with point (0,300) but this can't be can it since that means there are zero children?
 
Can you please post the graph so i can check to see what the result is? Why can't x=0??
 
amathproblemthatneedsolve said:
"there cannot be more than three children’s tickets sold for every adult ticket sold."
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doesn't x represent children tickets? Yes, that is how YOU defined x.
Well if x=0, then I would think that there will NOT be more than three children’s tickets sold for every adult ticket sold. Do you see this?
 
Suppose x=0.
If y=7, then y>3x, ie 7>0
If y=2, the y>3x, ie 2>0.
Note that y is never negative, so if x=o then it is always true that y>3x.
Now do you understand?
 
Well if this is the case am I suppose to use (0,300) ? P=7.5x+7.5y =2250 is the new profit and there are only 300 adult tickets sold. I feel like I did something wrong?
 
You failed to supply me with the graph as I requested so I do not know if you are correct. If you have the correct feasible region, found the correct corner points, computed P for each corner point and found that the highest P value was at (0, 300), then you have the correct solution.

Where do you think that you made a mistake?

Basically you want to know what the largest x+y can be (that comes from P=.75x+.75y)


1)x≥0
2)y≥0
3)x+y≤300 ----- this allows x and y to be as much as 300
4)12x+3y≥900----This says that 4x+y ≥300, so it is better to use more y's than x's
5)4x+8y≥1000---x+2y≥250, this does not go against x to be as much as 75 and y to be as much as 300

So up to here it is best to use y=300.
6)y≥3x----this inequality is fine if y=300 and x=0.
 
Jomo said:
But it only shows the lines. Your constraints are not lines but rather inequalities. In the graph in post 9 it does not show the feasible regions.
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OK, so what do I need to do to get my answer? (note its P=7.5x+7.5y) are my feasible points wrong? (0,300) , (75,225), (42., 128.)?
 
amathproblemthatneedsolve said:
OK, so what do I need to do to get my answer? note its P=7.5x+7.5y now are my feasible points wrong? (0,300) , (75,225), (42., 128.)?
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I can't be any clearer. I am not going to draw the graph. You draw the graph, shade away the bad points and then I will be able to see the feasible set and can see if (0, 300) is the solution. Although in my previous post (#16), I did verify your result. But fine, as a beginner you want to do it the typical way which you learned.
 
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