Original question

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 $f = (x + 2)^{-2}$

So far I've looked here and determined the equation for the height is $$h = \frac{-2}{(x+2)^3}\frac{1}{4}a$$ but when I try setting it to zero in the other answer the x doesn't cancel. I also tried wolframalpha so I'm pretty sure I did not simplify wrong. Any help would be appreciated.

  • $\begingroup$ The area of the triangle is proportional to the product of the $x$- and $y$-intercepts of the tangent line. $\endgroup$ – amd Nov 16 '17 at 1:41

We must fix a parameter. It's convenient to have that as the $x$ coordinate of the point on the curve where the tangent touches the curve. Let this be $a$.

Slope of the tangent at that point = $\displaystyle -\frac{2}{(a+2)^3}$

Furthermore, the tangent passes through the point $\displaystyle (a,\frac{1}{(a+2)^2})$

So the equation of the tangent line is given by:

$\displaystyle y - \frac{1}{(a+2)^2} = -\frac{2}{(a+2)^3}(x-a)$

Set $y=0$ for $x$-intercept $x_0$ of the tangent line:

$\displaystyle - \frac{1}{(a+2)^2} = -\frac{2}{(a+2)^3}(x_0-a)$

Then set $x=0$ for $y$-intercept $y_0$ of the tangent line:

$\displaystyle y_0 - \frac{1}{(a+2)^2} = -\frac{2}{(a+2)^3}(-a)$

Solve for $x_0$ and $y_0$. The area of the triangle is $\frac 12x_0y_0$. Maximise that expression relative to $a$.


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