To solve this problem, we need to analyze the function $f(x, y) = \sqrt{4x + 3y}$ and find its gradient at the point $(-2, 4)$. The problem also asks us to sketch this gradient along with the level curve at this point and choose the correct graph.

Step 1: Compute the Gradient

The gradient vector $\nabla f(x, y)$ of a function $f(x, y)$ is given by: $\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).$

For $f(x, y) = \sqrt{4x + 3y}$, let's calculate the partial derivatives.

1. Partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{4x + 3y}} \cdot 4 = \frac{2}{\sqrt{4x + 3y}}.$

2. Partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{4x + 3y}} \cdot 3 = \frac{3}{2\sqrt{4x + 3y}}.$

Step 2: Evaluate the Gradient at $(-2, 4)$

Substitute $x = -2$ and $y = 4$ into the partial derivatives:

1. First, calculate $4x + 3y$ at $(-2, 4)$: $4(-2) + 3(4) = -8 + 12 = 4.$

2. Now substitute into each component:

   • For $\frac{\partial f}{\partial x}$: $\frac{\partial f}{\partial x} = \frac{2}{\sqrt{4}} = \frac{2}{2} = 1.$
   • For $\frac{\partial f}{\partial y}$: $\frac{\partial f}{\partial y} = \frac{3}{2\sqrt{4}} = \frac{3}{4}.$

So, the gradient at $(-2, 4)$ is: $\nabla f(-2, 4) = \left( 1, \frac{3}{4} \right).$

Step 3: Analyze the Level Curve and Gradient Direction

The gradient vector $(1, \frac{3}{4})$ represents the direction of the steepest ascent from the point $(-2, 4)$. This vector points in the positive $x$- and $y$-directions. In the graphs, this would be represented by an arrow originating at $(-2, 4)$ and pointing in the first quadrant.

Step 4: Identify the Correct Graph

The correct graph will show:

• A level curve for $f(x, y) = \sqrt{4x + 3y}$ passing through $(-2, 4)$.
• A gradient vector at $(-2, 4)$ pointing in the direction $(1, \frac{3}{4})$, i.e., towards the upper right quadrant.

Based on this analysis, the correct graph should display the gradient vector pointing diagonally up and to the right from $(-2, 4)$, which corresponds to option B in the provided graphs.

Would you like more details on any part of this solution? Here are some related questions for further exploration:

1. How does the gradient vector relate to the direction of steepest ascent?
2. What is a level curve, and how is it useful in understanding multivariable functions?
3. Can the gradient be zero? If so, under what circumstances?
4. How would you find the directional derivative along a given direction for this function?
5. What are some applications of gradient vectors in real-world scenarios?

Tip: In multivariable calculus, the gradient vector at a point on a surface points perpendicular to the level curve at that point, which helps in visualizing the direction of change.