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.
- 1 decade agoFavorite 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
- PuggyLv 71 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
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 71 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 - 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) !
- yljacktt81Lv 51 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
Now the x-intercept is value of
x when y=0, so
So area is (10-6m)((6m-10)/m)*.5=
Now take derviative and set to 0, so
Now m = -1.666.since it cuts through
Now equation is y=-1.666x+b, and plug in (6,10) to
find b, so 10=(-1.666*6)+b, so
So equation is y=-1.6667x+20, but check calculations.