Find the maximum area of a triangle formed in the first quadrant by the
x-axis,
7-axis and a tangent line to the graph of y=(x+7)^-2

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.

Explore how to find the maximum area of a triangle in the first quadrant formed by the x-axis, y-axis, and a tangent line to the curve y=(x+7)^-2. Learn step-by-step with calculus methods and optimization techniques.

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