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:

Always simplify algebraic fractions before taking derivatives!
Rewriting \(\frac{82-10x}{10-x}\) as \(10 - \frac{18}{10-x}\) makes finding the derivative \(f'(x)\) much faster and less prone to quotient-rule errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We want to maximize the total revenue from selling two types of dolls.
Let \(x\) (hundreds) be the number of flopsy dolls and \(y\) (hundreds) be the number of mopsy dolls.
The relation between \(y\) and \(x\) is given by the production constraint: \(y = \frac{82-10x}{10-x}\) for \(0 \le x \le 8\).
The price of a mopsy doll is twice that of a flopsy doll.
Key Formula or Approach:
Let the price of a flopsy doll be \(P\) (per hundred). Then the price of a mopsy doll is \(2P\).
The total revenue \(R\) is:
\[ R = P \cdot x + 2P \cdot y = P(x + 2y) \]
To maximize the revenue \(R\), we need to maximize the function:
\[ f(x) = x + 2y = x + 2\left(\frac{82-10x}{10-x}\right) \]

Step 2: Detailed Explanation:

• Simplify the expression for \(y\):
\[ y = \frac{82-10x}{10-x} = \frac{10(10-x) - 18}{10-x} = 10 - \frac{18}{10-x} \]

• Substitute this simplified \(y\) into the objective function \(f(x)\):
\[ f(x) = x + 2\left(10 - \frac{18}{10-x}\right) = x + 20 - \frac{36}{10-x} \]

• Differentiate \(f(x)\) with respect to \(x\):
\[ f'(x) = 1 - 36(-1)(10-x)^{-2}(-1) = 1 - \frac{36}{(10-x)^2} \]

• Set \(f'(x) = 0\) to find the critical points:
\[ 1 - \frac{36}{(10-x)^2} = 0 \implies (10-x)^2 = 36 \] \[ 10-x = \pm 6 \]
Since \(0 \le x \le 8\):
\[ 10-x = 6 \implies x = 4 \]
(The other root \(10-x = -6 \implies x = 16\) is outside the domain).

• Confirm the maximum using the second derivative test:
\[ f''(x) = \frac{d}{dx}\left(1 - 36(10-x)^{-2}\right) = -72(10-x)^{-3} \] At \(x = 4\):
\[ f''(4) = -\frac{72}{(10-4)^3} = -\frac{72}{216} = -\frac{1}{3} < 0 \] Since the second derivative is negative, \(x = 4\) yields a local maximum.

• Calculate the corresponding value of \(y\) at \(x = 4\):
\[ y = 10 - \frac{18}{10-4} = 10 - \frac{18}{6} = 10 - 3 = 7 \]

• Convert \(x\) and \(y\) back from hundreds to units:
- Flopsy: \(4 \times 100 = 400\) dolls.
- Mopsy: \(7 \times 100 = 700\) dolls.

Step 3: Final Answer:
The revenue is maximized when the manufacturer produces 400 flopsy dolls and 700 mopsy dolls.