Question

A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

• A. 5
• B. 75
• C. 15
• D. 70

Hint:

Here is an elegant algebra shortcut using the AM-GM inequality concepts: when splitting a constant sum $S$ into parts to maximize a product of powers like $x^1 \cdot y^3$, the optimal values are always directly proportional to their exponents! Thus, split 20 into the ratio $1:3$. The parts are $\frac{1}{4} \times 20 = 5$ and $\frac{3}{4} \times 20 = 15$. Their product is simply $15 \times 5 = 75$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A wire of fixed total length (20 units) is split into two fragments. We need to find the configuration that maximizes the product of one part and the cube of the other part, and then calculate the simple product of those two split values.

Step 2: Key Formula or Approach:
This is an optimization problem that can be solved using the first derivative test from differential calculus. Let the first part be $x$. Since the total sum of the parts is 20, the second part will be $(20 - x)$. We construct our function $f(x)$ representing the product of one part and the cube of the other: $$ f(x) = (20 - x)x^3 = 20x^3 - x^4 $$ To maximize this function, we set its first derivative to zero ($f'(x) = 0$) to find the critical points.

Step 3: Detailed Explanation: Let's find the first derivative of $f(x)$ with respect to $x$: $$ f'(x) = \frac{d}{dx}(20x^3 - x^4) = 60x^2 - 4x^3 $$ Setting $f'(x) = 0$ to find the critical numbers: $$ 60x^2 - 4x^3 = 0 $$ Factoring out $4x^2$: $$ 4x^2(15 - x) = 0 $$ This yields two possible values for $x$: $x = 0$ or $x = 15$. Since a part length of 0 would drop the product to zero, $x = 15$ is our optimal candidate. Let's verify using the second derivative test: $$ f''(x) = \frac{d}{dx}(60x^2 - 4x^3) = 120x - 12x^2 $$ Substituting $x = 15$: $$ f''(15) = 120(15) - 12(15)^2 = 1800 - 12(225) = 1800 - 2700 = -900 $$ Since $f''(15) < 0$, the function achieves a local maximum at $x = 15$. Now we can determine the lengths of both parts:

• First part: $x = 15$

• Second part: $20 - x = 20 - 15 = 5$
The question asks for the product of these two parts: $$ \text{Product} = 15 \times 5 = 75 $$

Step 4: Final Answer: The product of the two parts when the condition is maximized is 75, which matches option (B).