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 $$f=(x+6{)}^{-2}$$

Gabriella Hensley

Beginner2022-10-01Added 6 answers

Solution:

$$y=(x+6{)}^{-2}\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=-2(x+6{)}^{-3}=\frac{-2}{(x+6{)}^{3}}$$

At x=a

$$\frac{dy}{dx}=\frac{2}{(a+6{)}^{3}}$$

The equation of tangent at x=a is

$$y-\frac{1}{(a+6{)}^{2}}=\frac{-2}{(a+6{)}^{3}}(x-a)\phantom{\rule{0ex}{0ex}}\Rightarrow y\frac{(a+6{)}^{2}-1}{(a+6{)}^{2}}=\frac{-2(x-a)}{(a+6{)}^{3}}\phantom{\rule{0ex}{0ex}}y(a+6{)}^{3}-(a+6)=-2(x-a)\phantom{\rule{0ex}{0ex}}y(a+6{)}^{3}+2(x-a)-(a+6)=0$$

Set x=0 for y intercept

$$y(a+6{)}^{3}+2(x-a)-(a+6)=0\phantom{\rule{0ex}{0ex}}y=\frac{3a+6}{(a+6{)}^{3}}$$

Set y=0 for x-intercept

$$y(a+6{)}^{3}+2(x-a)-(a+6)=0\phantom{\rule{0ex}{0ex}}2(x-a)-(a+6)=0\phantom{\rule{0ex}{0ex}}x=\frac{3a+6}{2}$$

Now,

Area of triangle $$(A)=\frac{1}{2}xy\phantom{\rule{0ex}{0ex}}=\frac{1}{2}(\frac{3a+6}{2})(\frac{3a+6}{(a+6{)}^{3}})$$

To maximize this area we have to differentiate with respect to a

$$\frac{dA}{da}=\frac{d}{da}(\frac{1}{4}\frac{(3a+6{)}^{2}}{(a+6{)}^{3}})\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\cdot \frac{6(3a+6)(a+6{)}^{3}-3(a+6{)}^{2}(3a+6{)}^{2}}{((a+6{)}^{3}{)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3(3a+6)(-a+6)}{4(a+6{)}^{4}}\phantom{\rule{0ex}{0ex}}\frac{dA}{da}=0\phantom{\rule{0ex}{0ex}}3(3a+6)(-a+6)=0$$

which gives a=-2,6

when a=-2, A=0

when a=6, a=1/12

The maximum area $$=\frac{1}{12}$$

$$y=(x+6{)}^{-2}\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=-2(x+6{)}^{-3}=\frac{-2}{(x+6{)}^{3}}$$

At x=a

$$\frac{dy}{dx}=\frac{2}{(a+6{)}^{3}}$$

The equation of tangent at x=a is

$$y-\frac{1}{(a+6{)}^{2}}=\frac{-2}{(a+6{)}^{3}}(x-a)\phantom{\rule{0ex}{0ex}}\Rightarrow y\frac{(a+6{)}^{2}-1}{(a+6{)}^{2}}=\frac{-2(x-a)}{(a+6{)}^{3}}\phantom{\rule{0ex}{0ex}}y(a+6{)}^{3}-(a+6)=-2(x-a)\phantom{\rule{0ex}{0ex}}y(a+6{)}^{3}+2(x-a)-(a+6)=0$$

Set x=0 for y intercept

$$y(a+6{)}^{3}+2(x-a)-(a+6)=0\phantom{\rule{0ex}{0ex}}y=\frac{3a+6}{(a+6{)}^{3}}$$

Set y=0 for x-intercept

$$y(a+6{)}^{3}+2(x-a)-(a+6)=0\phantom{\rule{0ex}{0ex}}2(x-a)-(a+6)=0\phantom{\rule{0ex}{0ex}}x=\frac{3a+6}{2}$$

Now,

Area of triangle $$(A)=\frac{1}{2}xy\phantom{\rule{0ex}{0ex}}=\frac{1}{2}(\frac{3a+6}{2})(\frac{3a+6}{(a+6{)}^{3}})$$

To maximize this area we have to differentiate with respect to a

$$\frac{dA}{da}=\frac{d}{da}(\frac{1}{4}\frac{(3a+6{)}^{2}}{(a+6{)}^{3}})\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\cdot \frac{6(3a+6)(a+6{)}^{3}-3(a+6{)}^{2}(3a+6{)}^{2}}{((a+6{)}^{3}{)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3(3a+6)(-a+6)}{4(a+6{)}^{4}}\phantom{\rule{0ex}{0ex}}\frac{dA}{da}=0\phantom{\rule{0ex}{0ex}}3(3a+6)(-a+6)=0$$

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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