Find two nonnegative numbers x and y for which x+3y=30, such that (x^2)y is maximized

To solve this optimization problem, we need to find the values of x and y such that x + 3y = 30 and the expression (x^2)y is maximized.

### Step 1: Express y in terms of x
Given x + 3y = 30 , we have:
y = \frac{30 - x}{3}

### Step 2: Substitute y into the expression to maximize
Substitute y into (x^2)y :
f(x) = \frac{x^2(30 - x)}{3}

### Step 3: Differentiate and find critical points
Differentiate f(x) to find the critical points:
f'(x) = 20x - x^2
Setting derivative to zero:
20x - x^2 = 0
This gives x = 0 or x = 20 .

### Step 4: Evaluate f(x) at critical points and boundary points
Evaluate at x = 0, 20, 30 :
- For x = 0 : f(0) = 0
- For x = 20 : f(20) = \frac{4000}{9} \approx 444.44
- For x = 30 : f(30) = 0

### Step 5: Determine the maximum value
The maximum value is \frac{4000}{9} at x = 20 .

### Answer:
The maximum value of (x^2)y is approximately 444.44 when x = 20 and y = \frac{10}{3} .