January 17, 2017

Homework Help: Calculus

Posted by Kristal on Sunday, November 12, 2006 at 7:25pm.

I asked for help on a question yesterday and I got really good help and I was wondering if someone could help with this problem too.

This is the question:
Find an equation of the line through the point (1,3) that cuts off the least area from the first quadrant.

My work on the problem so far:
I know that it will be in the first quadrant and that the result will form a right triangle.
I have written the equation y=-mx+b because I do know that the line will be decreasing.
Besides that, I do not know what to do.
Can someone help me please?

If you take the equation for the line to be:

y=-mx+b (1)

then you have to require that if x = 1, y = 3. So, you insert these figures in Eq. (1) and you get:

3 = -m + b (2)

This equation does not fix m and b but relates the two. This is what you would expect, because you can draw a line through the point (1,3) in an infinite number of ways. You then calculate where the line will intersect the x-axis. At that point y=0, so:

-mx + b =0 -->

x = b/m

And the line intersects the Y-axis when x = 0, so the y-coordinate at that point is b.

This maens that the right traingle as its three corners at:

(0,0), (0,b) and (b/m,0)

The area of the triangle is thus:

A = 1/2 b^2/m (3)

If you use Eq. (2) to express b in terms of m:

b = 3 + m and insert that in (3), you find:

A = 1/2 (3 + m)^2/m =

1/2 (9/m + 6 + m)

If you equate the derivative of this to zero you find:

m = 3.

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