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28 September, 19:00

Given that x2+y2=16x-26y-133, what is the biggest value that 6x-8y can have?

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  1. 28 September, 20:35
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    Note that x^2+y^2=16x-26y-133 represents a circle.

    x^2 - 16x + y^2 + 26y = 133

    x^2 - 16x + 64 - 64 + y^2 + 26y + 169 - 169 = 133 (completing the square)

    Then (x-8) ^2 - 64 + (y+13) ^2 - 169 = 133

    Simplifying, (x-8) ^2 + (y+13) ^2 = 133 + 169 + 64 = 366

    This circle acts as a constraint on the value of 6x-8y. Assume that x and y are both on the circle. Just supposing that x = 10, find y:

    (10-8) ^2 + (y+13) ^2 = 366, or 4 + (y+13) ^2 = 366, or (y+13) ^2 = 362

    This is a quadratic equation that could be solved for y, and the result (s) could be subst. into the expression 6x-8y.

    If you were to repeat this exercise several times, for different values of x, you'd come up with various values of 6x-8y and in that way approach (if not find) a definite answer to " Given that x2+y2=16x-26y-133, what is the biggest value that 6x-8y can have?"

    Hope someone else can come up with a more elegant approach.
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