The marginal cost of a manufacture is given by:
`MC=(250x)/sqrt(x^2+40)`
where 'x' is the number of units (in thousands) produced. If 'x' increases from 3,000 to 9,000 units, find the total increase in cost.

Answer

To find the total increase in cost when the number of units produced increases from 3,000 to 9,000 units, we will follow these steps:

### Step 1: Understand the Marginal Cost Function
The marginal cost (MC) is given by the formula:
\[ MC = \frac{250x}{\sqrt{x^2 + 40}} \]
where \(x\) is the number of units produced in thousands.

### Step 2: Set Up the Integral
To find the total increase in cost when \(x\) increases from 3 to 9 (since 3,000 units is 3 and 9,000 units is 9 in thousands), we need to compute the integral of the marginal cost function from 3 to 9:
\[ \Delta C = \int_{3}^{9} MC \, dx = \int_{3}^{9} \frac{250x}{\sqrt{x^2 + 40}} \, dx \]

### Step 3: Use Substitution
Let’s use the substitution method. Set:
\[ t = x^2 + 40 \implies dt = 2x \, dx \implies dx = \frac{dt}{2x} \]
When \(x = 3\), \(t = 3^2 + 40 = 49\).
When \(x = 9\), \(t = 9^2 + 40 = 121\).

Now, substituting \(x\) and \(dx\) in the integral:
\[ \Delta C = \int_{49}^{121} \frac{250x}{\sqrt{t}} \cdot \frac{dt}{2x} = \int_{49}^{121} \frac{125}{\sqrt{t}} \, dt \]

### Step 4: Evaluate the Integral
Now we can evaluate the integral:
\[ \Delta C = 125 \int_{49}^{121} t^{-1/2} \, dt \]
The integral of \(t^{-1/2}\) is:
\[ \int t^{-1/2} \, dt = 2t^{1/2} \]
So, we have:
\[ \Delta C = 125 \cdot 2 \left[ t^{1/2} \right]_{49}^{121} = 250 \left[ \sqrt{121} - \sqrt{49} \right] \]
Calculating the square roots:
\[ \Delta C = 250 \left[ 11 - 7 \right] = 250 \cdot 4 = 1000 \]

### Step 5: Conclusion
The total increase in cost when the production increases from 3,000 to 9,000 units is:
\[ \Delta C = 1000 \]