SOLUTION: Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis and a tangent line to the graph of f=(x+8)^(−2) Area =

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Question 1183670: Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis and a tangent line to the graph of f=(x+8)^(−2)
Area =
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
The triangle with maximum area is a right triangle in the first quadrant,
because its legs are the values of the intercepts.

f%28x%29=%28x%2B8%29%5E%28-2%29

%22f%27%28x%29%22=-2%28x%2B8%29%5E%28-3%29

The point on the graph where x=k is %28matrix%281%2C3%2Ck%2C%22%2C%22%2C%28k%2B8%29%5E%28-2%29%29%29

The slope of a line tangent to the graph at the point where x=k is -2%28k%2B8%29%5E%28-3%29

The equation through that point with that slope is

y-%28k%2B8%29%5E%28-2%29=%28-2%28k%2B8%29%5E%28-3%29%29%28x-k%29

The x-coordinate of that line is %28matrix%281%2C3%2C%283k%2B8%29%2F2%2C%22%2C%22%2C0%29%29
The y-coordinate of that line is %28matrix%281%2C3%2C0%2C%22%2C%22%2C%283k%2B8%29%2F%28k%2B8%29%5E3%29%29

So %283k%2B8%29%2F2 and %283k%2B8%29%2F%28k%2B8%29%5E3 are the legs of the right triangle we wish
to maximize the area of.

The area of the triangle is

A=expr%281%2F2%29%28%283k%2B8%29%2F2%29%28%283k%2B8%29%2F%28k%2B8%29%5E3%29

A=expr%281%2F4%29%28%283k%2B8%29%5E2%2F%28k%2B8%29%5E3%29

%28dA%29%2F%28dk%29=+-%283%28k+-+8%29%283k+%2B+8%29%5E%22%22%29%2F%284%28k+%2B+8%29%5E4%29

Since the triangle is in the first quadrant, the only feasible value of k
for which that derivative is zero is k=8. So we substitute k=8,

A=expr%281%2F4%29%28%283%2A8%2B8%29%5E2%2F%288%2B8%29%5E3%29

That works out to be 1%2F16

Edwin