Calculus optimization problem?

Find an equation of the line through the point (6,10) that cuts off the least area from the first quadrant.

I know how to do most optimization problems, but I don't know what equations I should use to approach this problem.

4 Answers

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  • 1 decade ago
    Favorite Answer

    Good to hear that you could do optimization problems.

    Back to your problem. Let say m is the slope of the straight line that goes thru the point (6,10). The equation of the line would be y = m(x-6) + 10 or y = mx - 6m + 10.

    The straight line intersects the two coordinate axes at (0, -6m+10) and ((6m-10)/m, 0) consecutively.

    Therefore, the area of the triangle that the line cut off from the first quadrant is A(m) = (1/2)*( -6m+10)*(6m-10)/m = -(6m-10)^2/(2m)

    Then using the first derivative test (or a graphing calculator), you can determine that the area would be smallest if m = - 5/3,

    Therefore, the equation of that line must be

    y = -5x/3 +20

  • Puggy
    Lv 7
    1 decade ago

    The equation of a line is

    y = mx + b

    Which means at (6, 10), the equation is

    10 = m(6) + b, or

    10 = 6m + b

    or

    b = 10 - 6m

    The x-intercept of this line represents the base of the triangle.

    The x-intercept is found by making y = 0, so

    y = mx + b

    0 = mx + b

    -b = mx

    x = -b/m <==== the base.

    The y-intercept represents the height of the triangle, found by making x = 0.

    y = mx + b

    y = b <====== the height

    The formula of the area of the triangle is (1/2)(base)(height). Therefore,

    A = (1/2)(-b/m)(b)

    A = (1/2)(-b^2 / m)

    A = (-1/2)(b^2/m)

    But we know that b = 10 - 6m, so

    A = (-1/2)( [10 - 6m]^2 / m )

    So this will be our area formula, A(m)

    A(m) = (-1/2)( [10 - 6m]^2 / m )

    We want to minimize the area, so we take the derivative as normal.

    A'(m) = (-1/2) [ 2(10 - 6m)(-6)m - (10 - 6m)^2 (1) ] / [ m^2 ]

    A'(m) = (-1/2) [ (-12m)(10 - 6m) - (10 - 6m)^2 ] / [ m^2 ]

    A'(m) = (-1/2) [ (10 - 6m)(-12m - (10 - 6m)) ] / [ m^2 ]

    A'(m) = (-1/2) [ (10 - 6m)(-12m - 10 + 6m) ] / [ m^2 ]

    A'(m) = (-1/2) [ (10 - 6m)(-6m - 10) ] / [ m^2 ]

    Now, we make A'(m) = 0.

    0 = (-1/2) [ (10 - 6m)(-6m - 10) ] / [ m^2 ]

    Critical values: 5/3, -5/3

    We can reject 5/3 since it is a positive slope (the only way to form triangles from the first quadrant is if the slope is negative). That leaves -5/3, so this tells us the minimum area occurs when the slope m = -5/3.

    To find the equation of the line, all we have to do is use high school methods to find the equation of the line with slope m = = -5/3 and through (6, 10).

    (y2 - y1)/(x2 - x1) = -5/3

    (y - 10)/(x - 6) = -5/3

    y - 10 = (-5/3)(x - 6)

    y - 10 = (-5/3)x + 10

    y = (-5/3)x + 20

  • ??????
    Lv 7
    1 decade ago

    y-10 = m(x-6)

    is the equation of all lines through (6,10)

    the are cut off is a triangle

    to know the height and base of this triangle

    we have to calculate the intersections with

    x-axis and y-axis.

    x=0 => y=10-6m

    y=0 => x=6-10/m

    so the area is (10-6m)(6-10/m)/2

    = (60+60-36m-100/m)/2

    = 60 - 18m - 50/m

    We have to take the derivative and put it zero

    to find a minimum

    = -18 + 50/m² = 0

    => 50/m² = 18

    => m² = 50/18 = 25/9

    => m = +- 5/3

    to see if it's a minimum, derive again

    -100/m³ > 0 if m<0 so we have to take the minus sign

    and the equation is

    y-10 = -5/3 (x-6) !

  • 1 decade ago

    Now using y=mx+b form, we

    know that area is b*(x-intercept)*.5.

    Now we know that 10=6m+b or

    b=10-6m.

    So y=mx+(10-6m).

    Now the x-intercept is value of

    x when y=0, so

    0=mx+10-6m, or

    (6m-10)/m=x.

    So area is (10-6m)((6m-10)/m)*.5=

    (10-6m)(6-(10/m))*.5=

    30-(50/m)-(18m)+30.

    Now take derviative and set to 0, so

    (50/m^2)-18=0, so

    50/m^2=18, or

    m^2=50/18, or

    m=+/-1.666...

    Now m = -1.666.since it cuts through

    1st quadrant.

    Now equation is y=-1.666x+b, and plug in (6,10) to

    find b, so 10=(-1.666*6)+b, so

    10+10=b=20.

    So equation is y=-1.6667x+20, but check calculations.

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