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: Set up the function.
Let one part be $x$, the other $20 - x$. Maximise $f(x) = (20 - x)x^3$.

Step 2: Differentiate and solve.
$f'(x) = 60x^2 - 4x^3 = 4x^2(15 - x)$. Setting this to zero gives $x = 15$, which is a maximum.

Step 3: Find the product.
Parts are 15 and 5, so the product is $15 \times 5 = 75$.
\[ \boxed{75,\ \text{option 2}} \]