Question

A toy manufacturer produces an inexpensive doll (flopsy) and an expensive doll (mopsy) in units of x hundreds and y hundreds respectively. Suppose it is possible to produce the doll in such a way that y = \(\frac{82-10x}{10-x}\) , 0 \(\le\) x \(\le\) 8 and that the company receives twice as much for selling a mopsy doll as for selling a flopsy doll. The level of production for both flopsy and mopsy dolls for which the total revenue derived from selling these dolls is maximized is given by :

• A. Flopsy: 40 and Mopsy: 70
• B. Flopsy: 700 and Mopsy: 400
• C. Flopsy: 400 and Mopsy: 700
• D. Flopsy: 70 and Mopsy: 40

Hint:

To save time during the exam:
Note that \(x = 4\) is the only critical point in the range \([0, 8]\).
Since \(x = 4\) represents \(400\) units, only Option (C) matches this value.
This allows you to find the correct answer without calculating \(y\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is an optimization problem where we need to find the production levels that maximize the total revenue.
We can formulate the total revenue as a function of one variable, \(x\), and find its maximum using differentiation.
Key Formula or Approach:
Let the price of a flopsy doll be \(P\).
Since the company receives twice as much for selling a mopsy doll, the price of a mopsy doll is \(2P\).
The total revenue \(R\) is:
\[ R = P \cdot (100x) + 2P \cdot (100y) = 100P(x + 2y) \]
To maximize \(R\), we must maximize the function:
\[ f(x) = x + 2y \]
Given the relation:
\[ y = \frac{82-10x}{10-x} \]

Step 2: Detailed Explanation:
Let us substitute the expression for \(y\) into \(f(x)\):
\[ f(x) = x + 2\left(\frac{82-10x}{10-x}\right) \]
First, simplify the term inside the parentheses:
\[ \frac{82-10x}{10-x} = \frac{10(10-x) - 18}{10-x} = 10 - \frac{18}{10-x} \]
Now, substitute this back into \(f(x)\):
\[ f(x) = x + 2\left(10 - \frac{18}{10-x}\right) = x + 20 - \frac{36}{10-x} \]
To find the critical points, differentiate \(f(x)\) with respect to \(x\):
\[ f'(x) = \frac{d}{dx}\left(x + 20 - 36(10-x)^{-1}\right) = 1 - 36(-1)(10-x)^{-2}(-1) \]
\[ f'(x) = 1 - \frac{36}{(10-x)^2} \]
Set the derivative equal to zero to find the critical points:
\[ 1 - \frac{36}{(10-x)^2} = 0 \implies (10-x)^2 = 36 \]
Taking the square root on both sides:
\[ 10 - x = \pm 6 \]
This gives two possible values for \(x\):
- \(10 - x = 6 \implies x = 4\)
- \(10 - x = -6 \implies x = 16\) (which is outside the given range \(0 \le x \le 8\))
Thus, our only valid critical point is \(x = 4\).
Let us verify if this point maximizes the function using the second derivative:
\[ f''(x) = \frac{d}{dx}\left(1 - 36(10-x)^{-2}\right) = -36(-2)(10-x)^{-3}(-1) = -\frac{72}{(10-x)^3} \]
For \(x = 4\):
\[ f''(4) = -\frac{72}{(10-4)^3} = -\frac{72}{6^3} < 0 \]
Since the second derivative is negative, \(x = 4\) is indeed a local maximum.
Now, calculate the corresponding value of \(y\) at \(x = 4\):
\[ y = 10 - \frac{18}{10-4} = 10 - \frac{18}{6} = 10 - 3 = 7 \]
Since \(x\) and \(y\) are in units of hundreds:
- Level of Flopsy production: \(4 \times 100 = 400\) dolls.
- Level of Mopsy production: \(7 \times 100 = 700\) dolls.
This matches Option (C).

Step 3: Final Answer:
The correct option is (C).