mocatgesex

2022-09-30

Find the maximum area of a triangle formed in the first quadrant by the xx-axis, yy-axis and a tangent line to the graph of $f=\left(x+6{\right)}^{-2}$

Gabriella Hensley

Solution:
$y=\left(x+6{\right)}^{-2}\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=-2\left(x+6{\right)}^{-3}=\frac{-2}{\left(x+6{\right)}^{3}}$
At x=a
$\frac{dy}{dx}=\frac{2}{\left(a+6{\right)}^{3}}$
The equation of tangent at x=a is
$y-\frac{1}{\left(a+6{\right)}^{2}}=\frac{-2}{\left(a+6{\right)}^{3}}\left(x-a\right)\phantom{\rule{0ex}{0ex}}⇒y\frac{\left(a+6{\right)}^{2}-1}{\left(a+6{\right)}^{2}}=\frac{-2\left(x-a\right)}{\left(a+6{\right)}^{3}}\phantom{\rule{0ex}{0ex}}y\left(a+6{\right)}^{3}-\left(a+6\right)=-2\left(x-a\right)\phantom{\rule{0ex}{0ex}}y\left(a+6{\right)}^{3}+2\left(x-a\right)-\left(a+6\right)=0$
Set x=0 for y intercept
$y\left(a+6{\right)}^{3}+2\left(x-a\right)-\left(a+6\right)=0\phantom{\rule{0ex}{0ex}}y=\frac{3a+6}{\left(a+6{\right)}^{3}}$
Set y=0 for x-intercept
$y\left(a+6{\right)}^{3}+2\left(x-a\right)-\left(a+6\right)=0\phantom{\rule{0ex}{0ex}}2\left(x-a\right)-\left(a+6\right)=0\phantom{\rule{0ex}{0ex}}x=\frac{3a+6}{2}$
Now,
Area of triangle $\left(A\right)=\frac{1}{2}xy\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\frac{3a+6}{2}\right)\left(\frac{3a+6}{\left(a+6{\right)}^{3}}\right)$
To maximize this area we have to differentiate with respect to a
$\frac{dA}{da}=\frac{d}{da}\left(\frac{1}{4}\frac{\left(3a+6{\right)}^{2}}{\left(a+6{\right)}^{3}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\cdot \frac{6\left(3a+6\right)\left(a+6{\right)}^{3}-3\left(a+6{\right)}^{2}\left(3a+6{\right)}^{2}}{\left(\left(a+6{\right)}^{3}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left(3a+6\right)\left(-a+6\right)}{4\left(a+6{\right)}^{4}}\phantom{\rule{0ex}{0ex}}\frac{dA}{da}=0\phantom{\rule{0ex}{0ex}}3\left(3a+6\right)\left(-a+6\right)=0$
which gives a=-2,6
when a=-2, A=0
when a=6, a=1/12
The maximum area $=\frac{1}{12}$

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