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Calculus/calculus optimization problem


A line through the point P(2,2) cuts the x and y axes at A and B, respectively. Find the minimum length of line segment AB.

Sorry about the fact that this a few days to answer,
but other matters gave me other stuff to do.

The equation of such a line would be y-2 = m(x-2).
As can be seen, (2,2) would be a point on that line.
Rewritten, this is y = mx - 2m + 2.

If it crosses the x-axis at x = A, the point is (A,0).
This makes the line pass through (2,2) and (A,0).
This makes the slope m = (2-0)/(2-A) = 2/(2-A).

Putting that back in the equation gives y = 2x/(2-A) - 2(2/(2-A)) + 2.
The last two terms combine to -4/(2-A) + (4-2A)/(2-A) = -2A/(2-A) = 2A/(A-2).
That makes the line be y = 2x/(2-A) + 2A/(A-2).

As seen from this, when x-0, y = 2A/(A-2).

It is known the the distance squared is D(A) = A +[2A/(A-2)].
That is, D(A) = A + 4A/(A-2).

Taking the derivative gives d(D)/dA = 2A - [8A(A-2) - 8A(A-2)]/(A-2)^4.
Cancelling an A-2 on the top and bottom of the 2nd term gives
d(D)/dA = 2A - [8A(A-2) - 8A]/(A-2).

Noting that 8A(A-2) = 8A-16A, we can subtract off 8A and get -16A.
This makes d(D)/dA = 2A - 16A/(A-2).

Solving 0 = 2A - 16A/(A-2) leads to adding 16A/(A-2) to both sides.
This gives 16A/(A-2) = 2A.
Multiplying both sides by (A-2)/(2A) gives 8 = (A-2).
Since 8 = 2, that means that 2 = A - 2, or A = 4.

As can be seen, if the line crosses the x-axis at A = 4, the slope is -1.
This means it crosses the y-axis at 4 as well.

The smallest area, then, is right triangle with both angles (besides the right angle) 45.  


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