A curve passes through the point (2,-8) and the slope of tangent at any point (x,y) is given by `x^(2)/2-6x`. The maximum ordinate on the curve is given by `lambda` then find `(3lambda`).

Answer

To solve the problem step-by-step, we will follow the instructions provided in the video transcript.

### Step 1: Understand the problem
We are given a curve that passes through the point (2, -8) and the slope of the tangent at any point (x, y) is given by the expression \( \frac{x^2}{2} - 6x \). We need to find the maximum ordinate (y-coordinate) on the curve, denoted as \( \lambda \), and then calculate \( 3\lambda \).

### Step 2: Set up the differential equation
The slope of the tangent at any point on the curve is given by:
\[ \frac{dy}{dx} = \frac{x^2}{2} - 6x \]

### Step 3: Integrate to find the function \( y \)
To find the equation of the curve, we need to integrate the expression for the slope:
\[ y = \int \left( \frac{x^2}{2} - 6x \right) dx \]
Calculating the integral:
\[ y = \frac{x^3}{6} - 3x^2 + C \]
where \( C \) is the constant of integration.

### Step 4: Use the point (2, -8) to find \( C \)
We know that the curve passes through the point (2, -8). We will substitute \( x = 2 \) and \( y = -8 \) into the equation to find \( C \):
\[ -8 = \frac{2^3}{6} - 3(2^2) + C \]
Calculating the terms:
\[ -8 = \frac{8}{6} - 3(4) + C \]
\[ -8 = \frac{4}{3} - 12 + C \]
\[ -8 + 12 = \frac{4}{3} + C \]
\[ 4 = \frac{4}{3} + C \]
Subtracting \( \frac{4}{3} \) from both sides:
\[ C = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3} \]

### Step 5: Write the complete equation of the curve
Now we can write the complete equation of the curve:
\[ y = \frac{x^3}{6} - 3x^2 + \frac{8}{3} \]

### Step 6: Find the maximum ordinate
To find the maximum ordinate, we need to differentiate \( y \) and set the derivative equal to zero:
\[ \frac{dy}{dx} = \frac{x^2}{2} - 6x \]
Setting the derivative to zero:
\[ \frac{x^2}{2} - 6x = 0 \]
Factoring out \( x \):
\[ x\left(\frac{x}{2} - 6\right) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad \frac{x}{2} = 6 \implies x = 12 \]

### Step 7: Determine the nature of the critical points
To determine whether these points are maxima or minima, we can find the second derivative:
\[ \frac{d^2y}{dx^2} = x - 6 \]
Evaluating the second derivative at the critical points:
- For \( x = 0 \):
\[ \frac{d^2y}{dx^2} = 0 - 6 = -6 \quad (\text{negative, hence maximum}) \]
- For \( x = 12 \):
\[ \frac{d^2y}{dx^2} = 12 - 6 = 6 \quad (\text{positive, hence minimum}) \]

### Step 8: Calculate the maximum ordinate
Now we calculate the maximum ordinate at \( x = 0 \):
\[ y(0) = \frac{0^3}{6} - 3(0^2) + \frac{8}{3} = \frac{8}{3} \]
Thus, the maximum ordinate \( \lambda = \frac{8}{3} \).

### Step 9: Calculate \( 3\lambda \)
Finally, we calculate \( 3\lambda \):
\[ 3\lambda = 3 \times \frac{8}{3} = 8 \]

### Final Answer
The value of \( 3\lambda \) is \( \boxed{8} \).