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Expert: - 11/7/2011


Question
The function, f(x)= 6(x+1)^-1 is graphed in the first quadrant and a tangent line is drawn at the point (a,b). The tangent line hits the axes at points P and Q. What point (a,b) should be chosen so that the area of the triangle PQO is minimized?

Please explain and show work. Thank you!

Answer
The slope is the derivative, f'(x).
Since f(x) = 6/(x+1), f'(x) = -6/(x+1)².
The point x0 that is chosen has the values (x0, 6/(x0 + 1)).
The slope of the line is -6/(x0 + 1)².

The points P and Q are found by finding where a line with this slop through the point given hits each of the axis.  If we take (f0 - 0)/(x0 - x) = m, f0, x0, and m are known, so solve for x.
This is where the curve crosses the x-axis.  If we take (f0 - y)/(x0 - 0) = m, this gives the slope.  Since f0, x0, and m are known, y can be solved for.

The area is xy/2 where x and y are the points just found from (x0,y0).
Now that the are is known in terms of x0, take x0 as the variable.
Put the area equation in terms of x0, and then find the derivative
of the area with respect to x0.

Set the derivative to 0, and that can be solved to find where x0 is to get the minimum area.

Since f0 and xo

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