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Denise
 5 years ago
How do I find the maximum profit....
0.8x^2+60x+120....
Denise
 5 years ago
How do I find the maximum profit.... 0.8x^2+60x+120....

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do you know what the graph of this equation is going to resemble?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it looks like an upside down "U".... or rather, a frown.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the highest point of this graph tells you the max amount it can produce; and we call that point of this graph the vertex.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if you know calculus you can also get it by taking the derivative and setting it equal to zero; which gives you the vertex anyway :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the x value of the vertex is given by the determinate of the quadratic formula: b/2a in this case that amounts to: 60/2(.8)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the max profit is then: .8(15/.4)^2 +60(15/.4) + 120 = max profits

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0either i miscalculated or its a bad equation to begin with :) I get: 1005

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its a bad equation :) http://www.wolframalpha.com/input/?i=0.8x%5E2%2B60x%2B120

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that should be a (0.8x^2) if anything

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0otherwise the max profits are infinite :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is no maximum. as stated the problem might be x^2? Is there a domain given for X? If this problem is stated correctly it will be a boundary for X.
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