The total cost function for a production is given by `C(x)=3/(4)x^(3)-7x+27`.
Find the number of units produced for which M.C. = A.C. {M.C. = Marginal Cost and A.C. = Average Cost}

Answer

To solve the problem, we will find the number of units produced for which the Marginal Cost (M.C.) equals the Average Cost (A.C.) using the given total cost function \( C(x) = \frac{3}{4}x^3 - 7x + 27 \).

### Step 1: Find the Marginal Cost (M.C.)
The Marginal Cost is the derivative of the total cost function \( C(x) \).

\[ C'(x) = \frac{d}{dx}\left(\frac{3}{4}x^3 - 7x + 27\right) \]

Using the power rule of differentiation:

\[ C'(x) = \frac{3}{4} \cdot 3x^2 - 7 = \frac{9}{4}x^2 - 7 \]

Thus, the Marginal Cost is:

\[ M.C. = \frac{9}{4}x^2 - 7 \]

### Step 2: Find the Average Cost (A.C.)
The Average Cost is given by the total cost divided by the number of units produced \( x \):

\[ A.C. = \frac{C(x)}{x} = \frac{\frac{3}{4}x^3 - 7x + 27}{x} \]

Simplifying this expression:

\[ A.C. = \frac{3}{4}x^2 - 7 + \frac{27}{x} \]

### Step 3: Set M.C. equal to A.C.
We need to find \( x \) such that:

\[ M.C. = A.C. \]

Substituting the expressions we found:

\[ \frac{9}{4}x^2 - 7 = \frac{3}{4}x^2 - 7 + \frac{27}{x} \]

### Step 4: Simplify the Equation
First, we can cancel \(-7\) from both sides:

\[ \frac{9}{4}x^2 = \frac{3}{4}x^2 + \frac{27}{x} \]

Next, multiply through by \( 4x \) to eliminate the fraction:

\[ 9x^3 = 3x^3 + 108 \]

### Step 5: Rearranging the Equation
Rearranging gives:

\[ 9x^3 - 3x^3 = 108 \]

\[ 6x^3 = 108 \]

### Step 6: Solve for \( x \)
Dividing both sides by 6:

\[ x^3 = 18 \]

Taking the cube root of both sides:

\[ x = \sqrt[3]{18} \]

Calculating the cube root gives approximately:

\[ x \approx 2.62 \]

### Step 7: Conclusion
Since we are looking for the number of units produced, we round \( x \) to the nearest whole number:

\[ x \approx 3 \]

Thus, the number of units produced for which Marginal Cost equals Average Cost is approximately **3 units**.

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