Question

On problem 2D-5* in Applications of Differentiation, I cannot find the solution on the answer key. Does anyone have a hint or do you know where to find the solution online? I seem to be missing something.

Answer 1

What's the question?

Answer 2

The question is: Find the triangle of smallest area in the half-plane to the right of the y axis, whose three sides are respectively segments of the x -axis, the diagonal y=x, and a line through (2,1).

Answer 3

Here's what I did:
Let us assume that the line that passes through 2,1 meets the x axis at x1,0

Since the line passes through 2,1 and x1,0 the equation is: y=-x/(x1-2)+(2/(x1-2) +1)
Now this line intersects y=x at x1/(x1-1)... that is A= (x1/(x1-1), x1/(x1-1))
Now area, a= 1/2(base)*altitude= 1/2*x1*(x1/(x1-1))

Now, remember that x1 is the variable, and you need the smallest triangle, so differentiate area wrt x1 and equate it to 0

da/dx1= d/dx1*(1/2*(x1^2)/x1-1)= 1/2 (d/dx1(x1+1)+d/dx1(1/(x1-1))=0
Solving it gives x1=2, now using it find the co-ordinates of A,B and get the area.

PS: Sorry for the late reply. :/

Answer 4

Thanks. It really helped.