If `x ^(2) + y^(2) - 14 x - 6y - 6=0,` then the maximum possible value of `3x + 4y` is (where (x,y) is a point on the given curve)

Options

A. 71
B. 72
C. 73
D. none of these

Answer

To find the maximum possible value of \(3x + 4y\) given the equation of the curve \(x^2 + y^2 - 14x - 6y - 6 = 0\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ x^2 + y^2 - 14x - 6y - 6 = 0 \]
We will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 14x = (x - 7)^2 - 49 \]

For \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x - 7)^2 - 49 + (y - 3)^2 - 9 - 6 = 0 \]
Simplifying this:
\[ (x - 7)^2 + (y - 3)^2 - 64 = 0 \]
Thus, we have:
\[ (x - 7)^2 + (y - 3)^2 = 64 \]
This represents a circle centered at \((7, 3)\) with a radius of \(8\).

### Step 2: Parameterize the points on the circle
We can express the coordinates \((x, y)\) on the circle using trigonometric functions:
\[ x = 7 + 8 \cos \theta \]
\[ y = 3 + 8 \sin \theta \]

### Step 3: Substitute into the expression \(3x + 4y\)
Now we substitute \(x\) and \(y\) into the expression \(3x + 4y\):
\[ 3x + 4y = 3(7 + 8 \cos \theta) + 4(3 + 8 \sin \theta) \]
Expanding this:
\[ = 21 + 24 \cos \theta + 12 + 32 \sin \theta \]
Combining like terms:
\[ = 33 + 24 \cos \theta + 32 \sin \theta \]

### Step 4: Find the maximum value
To maximize \(33 + 24 \cos \theta + 32 \sin \theta\), we can rewrite it in the form \(R \cos(\theta - \phi)\):
\[ R = \sqrt{24^2 + 32^2} = \sqrt{576 + 1024} = \sqrt{1600} = 40 \]
Now, we need to find \(\phi\) such that:
\[ \cos \phi = \frac{24}{40} = \frac{3}{5}, \quad \sin \phi = \frac{32}{40} = \frac{4}{5} \]

Thus, we can express:
\[ 24 \cos \theta + 32 \sin \theta = 40 \cos(\theta - \phi) \]
The maximum value of \(\cos\) is \(1\), so:
\[ \text{Maximum of } 33 + 40 \cos(\theta - \phi) = 33 + 40 = 73 \]

### Final Answer
The maximum possible value of \(3x + 4y\) is:
\[ \boxed{73} \]