To find the maximum area of a triangle formed by the x-axis, the y-axis, and a tangent line to the graph of $y = \frac{1}{(x+7)^2}$, we can follow these steps:

Step 1: Find the equation of the tangent line

Given the function $y = \frac{1}{(x+7)^2}$, let's first find its derivative to determine the slope of the tangent line at any point $(x_0, y_0)$.

The derivative $y'$ of the function is:

$y' = \frac{d}{dx} \left( \frac{1}{(x+7)^2} \right) = -2(x+7)^{-3} = -\frac{2}{(x+7)^3}$

So, the slope of the tangent line at any point $(x_0, y_0)$ is:

$m = -\frac{2}{(x_0+7)^3}$

The equation of the tangent line at this point can be written as:

$y - y_0 = m(x - x_0)$

Substituting $y_0 = \frac{1}{(x_0+7)^2}$ and $m = -\frac{2}{(x_0+7)^3}$, we have:

$y - \frac{1}{(x_0+7)^2} = -\frac{2}{(x_0+7)^3}(x - x_0)$

This simplifies to:

$y = -\frac{2}{(x_0+7)^3}x + \frac{2x_0 + 7}{(x_0+7)^3}$

Step 2: Find the x- and y-intercepts of the tangent line

The x-intercept occurs when $y = 0$:

$0 = -\frac{2}{(x_0+7)^3}x + \frac{2x_0 + 7}{(x_0+7)^3}$

$x = x_0 + 7$

The y-intercept occurs when $x = 0$:

$y = -\frac{2}{(x_0+7)^3}(0) + \frac{2x_0 + 7}{(x_0+7)^3} = \frac{2x_0 + 7}{(x_0+7)^3}$

Thus, the x-intercept is $x_0 + 7$, and the y-intercept is $\frac{2x_0 + 7}{(x_0+7)^3}$.

Step 3: Find the area of the triangle

The area $A$ of the triangle formed by the x-axis, the y-axis, and the tangent line is given by:

$A = \frac{1}{2} \times \text{x-intercept} \times \text{y-intercept}$

$A = \frac{1}{2} \times (x_0 + 7) \times \frac{2x_0 + 7}{(x_0+7)^3}$

This simplifies to:

$A = \frac{(2x_0 + 7)}{2(x_0+7)^2}$

Step 4: Maximize the area

To find the value of $x_0$ that maximizes the area, differentiate $A$ with respect to $x_0$ and set it equal to zero:

$\frac{dA}{dx_0} = \frac{d}{dx_0} \left( \frac{2x_0 + 7}{2(x_0+7)^2} \right)$

Using the quotient rule, we have:

$\frac{dA}{dx_0} = \frac{(2(x_0+7)^2) - (2x_0 + 7)(2(x_0+7))}{2(x_0+7)^4}$

Simplify the expression:

$\frac{dA}{dx_0} = \frac{2(x_0 + 7) - 2(2x_0 + 7)}{(x_0+7)^3} = \frac{2(x_0 + 7 - 2x_0 - 7)}{(x_0 + 7)^3} = \frac{-2x_0}{(x_0 + 7)^3}$

Setting $\frac{dA}{dx_0} = 0$, we have:

$-2x_0 = 0$

Thus, $x_0 = 0$.

Step 5: Calculate the maximum area

Substitute $x_0 = 0$ back into the expression for $A$:

$A_{\text{max}} = \frac{2(0) + 7}{2(0+7)^2} = \frac{7}{2 \times 49} = \frac{7}{98} = \frac{1}{14}$

So, the maximum area of the triangle is $\frac{1}{14}$.

Would you like further details, or have any questions?

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Here are some related questions for you:

1. What is the general method to find the tangent to any curve?
2. How does the area of the triangle change if we change the function $y = \frac{1}{(x+7)^2}$?
3. Can this problem be solved using Lagrange multipliers?
4. How does the area change if the tangent line was in a different quadrant?
5. What are the geometric interpretations of maxima and minima in calculus?
6. Can you generalize the result to other curves, e.g., $y = \frac{1}{(x+c)^2}$?
7. What if we had different axes, say $x = a$ and $y = b$, instead of the x- and y-axes?
8. How would the solution change if the function had a different power, e.g., $y = \frac{1}{(x+7)^3}$?

Tip: When dealing with optimization problems, always check if the function has any boundaries, as this can sometimes simplify finding the maximum or minimum values.