# 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.

Relevance
• 1 decade ago

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

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

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