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Math Help - Help with optimization problem?

  1. #1
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    Help with optimization problem?

    A straight line with a negative slope passes (4,5). Determine the line slope such that the area (in the first quadrant) between this line, x and y axis has a minimum.

    I'm totally stuck with this question and don't where to start from. anybody know how to solve this?
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  2. #2
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    Let f be the function describing the line. Since it passes through (4,5), f(x) = 5 - b(x-4) where -b (b>0) is the slope of the line. The shape formed in the first quadrant by this line and the axes is a triangle whose height is the y-intercept of the line and whose width is its x-intercept. You can compute the intercepts yourself to verify that they are \frac5b+4 and 5+4b. The area of the triangle is therefore \frac12(\frac5b+4)(5+4b) = 8b + \frac{25}{2b} + 20. Can you take it from here by setting the derivative to be zero? You should obtain b = 5/4, I think.
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  3. #3
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    Hi thanks for the assistance. I managed to find the height, but for my width i got X = -(5/b) - 4 instead.

    Using Y = bX + C, and taking y=0 and C = 5 +4m to find the x intercept,

    0=bX + 5 + 4b

    X=-(5/b) - 4

    May i know where i've gone wrong?
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  4. #4
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    Quote Originally Posted by toffeefan View Post
    Hi thanks for the assistance. I managed to find the height, but for my width i got X = -(5/b) - 4 instead.

    Using Y = bX + C, and taking y=0 and C = 5 +4m to find the x intercept,

    0=bX + 5 + 4b

    X=-(5/b) - 4

    May i know where i've gone wrong?
    First you are confusing "m" and "b". But mainly you have the wrong value for C. If Y= bX+ C and X= 4 when Y= 5, 5= 4b+ C so C= 5- 4C, not 5+ 4m or 5+ 4C.
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