A person wants to plant some trees in his community park. The local nursery has to perform this task. It charges the cost of planting trees by the formula :
`C(x )=x^(3)-45x^(2)+600x`,
where 'x' is the number of trees and C(x) is cost of planting 'x' trees in rupees. The local authority has imposed a restriction that it can plant 10 to 20 trees in one community park for a fair - districution. For how many trees should the person place the order so that he has to spend the least amount? How much is the least amount? Use Calculate to answer these questions.

Answer

To solve the problem of minimizing the cost of planting trees, we will follow these steps:

### Step 1: Define the Cost Function
The cost function given is:
\[ C(x) = x^3 - 45x^2 + 600x \]
where \( x \) is the number of trees.

### Step 2: Find the First Derivative
To find the critical points, we need to calculate the first derivative of the cost function:
\[ C'(x) = \frac{d}{dx}(x^3 - 45x^2 + 600x) \]
Using the power rule:
\[ C'(x) = 3x^2 - 90x + 600 \]

### Step 3: Set the First Derivative to Zero
To find the critical points, we set the first derivative equal to zero:
\[ 3x^2 - 90x + 600 = 0 \]
Dividing the entire equation by 3 simplifies it:
\[ x^2 - 30x + 200 = 0 \]

### Step 4: Solve the Quadratic Equation
We can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -30, c = 200 \):
\[ x = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 200}}{2 \cdot 1} \]
\[ x = \frac{30 \pm \sqrt{900 - 800}}{2} \]
\[ x = \frac{30 \pm \sqrt{100}}{2} \]
\[ x = \frac{30 \pm 10}{2} \]
This gives us two solutions:
\[ x = \frac{40}{2} = 20 \quad \text{and} \quad x = \frac{20}{2} = 10 \]

### Step 5: Determine the Nature of Critical Points
To determine whether these critical points are minima or maxima, we need to find the second derivative:
\[ C''(x) = \frac{d}{dx}(3x^2 - 90x + 600) \]
\[ C''(x) = 6x - 90 \]

Now we evaluate the second derivative at the critical points:

1. **At \( x = 10 \)**:
\[ C''(10) = 6(10) - 90 = 60 - 90 = -30 \] (which indicates a local maximum)

2. **At \( x = 20 \)**:
\[ C''(20) = 6(20) - 90 = 120 - 90 = 30 \] (which indicates a local minimum)

### Step 6: Calculate the Minimum Cost
Since \( x = 20 \) gives a local minimum, we substitute \( x = 20 \) back into the cost function to find the minimum cost:
\[ C(20) = 20^3 - 45(20^2) + 600(20) \]
Calculating each term:
\[ C(20) = 8000 - 45(400) + 12000 \]
\[ C(20) = 8000 - 18000 + 12000 \]
\[ C(20) = 8000 + 12000 - 18000 \]
\[ C(20) = 2000 \]

### Final Answer
The person should plant **20 trees** to spend the least amount, and the least amount is **2000 rupees**.

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