Find x Value for R-C?

Let R = 4x^2 + 8x and Let C = x^2 + 2x. For which value of x is R-C a maximum?

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  • Ray
    Lv 7
    6 years ago
    Best Answer

    R=4x^2+8x

    C=x^2+2x

    R-C= 3x^2+6x

    (R-C)'= 6x+6

    (R-C)"= 6

    Now (R-C)'=0 when x=-1

    Since (R-C)">0, its concave up at all times- slope increasing.

    So x=-1 is a minimum. There are no maximums for any x value.

  • 6 years ago

    R-C = 3x^2 + 6x = 3(x^2+2x) = 3(x^2+2x+1) - 3 = 3(x+1)^2 - 3.

    R-C has no maximum since x^2 can be as large as you like by picking x sufficiently large.

    R-C does have a minimum, since (x+1)^2 cannot be negative, the smallest it can be is 0 and that happens when x=-1. When x=-1 R-C = 3(x+1)^2 - 3 = 0 - 3 = -3 and that is the minimum value of R-C.

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