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 + 6)^{-2} ). give an exact answer.
area:

Explanation:

Step1: Find the derivative of the function

The function is \(y=(x + 6)^{-2}=\frac{1}{(x + 6)^{2}}\). Using the power - rule \((u^n)^\prime=nu^{n - 1}u^\prime\), where \(u=x + 6\) and \(n=-2\). Then \(y^\prime=-2(x + 6)^{-3}=-\frac{2}{(x + 6)^{3}}\).

Step2: Write the equation of the tangent line

The equation of the tangent line at the point \((a,\frac{1}{(a + 6)^{2}})\) is \(y-\frac{1}{(a + 6)^{2}}=-\frac{2}{(a + 6)^{3}}(x - a)\).
When \(x = 0\), \(y=\frac{1}{(a + 6)^{2}}+\frac{2a}{(a + 6)^{3}}=\frac{a + 6+2a}{(a + 6)^{3}}=\frac{3a + 6}{(a + 6)^{3}}\).
When \(y = 0\), \(0-\frac{1}{(a + 6)^{2}}=-\frac{2}{(a + 6)^{3}}(x - a)\), then \(x - a=\frac{(a + 6)}{2}\), so \(x=\frac{a + 6}{2}+a=\frac{3a + 6}{2}\).

Step3: Calculate the area of the triangle

The area of the triangle \(A=\frac{1}{2}\times x\times y\). Substitute \(x=\frac{3a + 6}{2}\) and \(y=\frac{3a + 6}{(a + 6)^{3}}\) into the area formula:
\(A=\frac{(3a + 6)^{2}}{4(a + 6)^{3}}\). Let \(t=a + 2\), then \(a=t - 2\) and \(A=\frac{9t^{2}}{4(t + 4)^{3}}\).
Differentiate \(A\) with respect to \(t\) using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 9t^{2}\), \(u^\prime=18t\), \(v = 4(t + 4)^{3}\), \(v^\prime=12(t + 4)^{2}\).
\(A^\prime=\frac{18t\times4(t + 4)^{3}-9t^{2}\times12(t + 4)^{2}}{16(t + 4)^{6}}=\frac{72t(t + 4)^{2}(t + 4-1.5t)}{16(t + 4)^{6}}=\frac{72t(t + 4)^{2}(4 - 0.5t)}{16(t + 4)^{6}}=\frac{9t(8 - t)}{2(t + 4)^{4}}\).

Step4: Find the critical points

Set \(A^\prime=0\). Then \(9t(8 - t)=0\), so \(t = 0\) or \(t = 8\). Since \(a>0\) (in the first - quadrant), when \(t = 8\) (i.e., \(a=6\)).

Step5: Calculate the maximum area

When \(a = 6\), \(x=\frac{3\times6+6}{2}=12\), \(y=\frac{3\times6 + 6}{(6 + 6)^{3}}=\frac{24}{1728}=\frac{1}{72}\).
\(A=\frac{1}{2}\times12\times\frac{1}{6}=\frac{1}{2}\times12\times\frac{1}{6}= \frac{1}{2}\times\frac{3\times6 + 6}{2}\times\frac{3\times6+6}{(6 + 6)^{3}}\)
\(A=\frac{1}{2}\times12\times\frac{1}{6}= \frac{3}{2}\)

Answer:

\(\frac{3}{2}\)