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Denise

  • 5 years ago

How do I find the maximum profit.... 0.8x^2+60x+120....

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  1. amistre64
    • 5 years ago
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    do you know what the graph of this equation is going to resemble?

  2. amistre64
    • 5 years ago
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    it looks like an upside down "U".... or rather, a frown.

  3. amistre64
    • 5 years ago
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    the highest point of this graph tells you the max amount it can produce; and we call that point of this graph the vertex.

  4. amistre64
    • 5 years ago
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    if you know calculus you can also get it by taking the derivative and setting it equal to zero; which gives you the vertex anyway :)

  5. amistre64
    • 5 years ago
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    the 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)

  6. amistre64
    • 5 years ago
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    the max profit is then: .8(-15/.4)^2 +60(-15/.4) + 120 = max profits

  7. amistre64
    • 5 years ago
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    either i miscalculated or its a bad equation to begin with :) I get: -1005

  8. amistre64
    • 5 years ago
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    its a bad equation :) http://www.wolframalpha.com/input/?i=0.8x%5E2%2B60x%2B120

  9. amistre64
    • 5 years ago
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    that should be a (-0.8x^2) if anything

  10. amistre64
    • 5 years ago
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    otherwise the max profits are infinite :)

  11. anonymous
    • 5 years ago
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    there 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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