Question

divide the number 84 into two parts such that the product of one part and the square of the other is maximum

Answer 1

let the two parts be x and y.
x+y = 84
⇒ y = (84-x)

You need to maximise the product of "one part" and "the square of other"
Let one part = (84-x)
square of other part = (x)² = x²
product, p = x² × (84-x) = 84x² - x³

$\frac{dp}{dx}=0\\ \\ \Rightarrow \frac{d}{dx}(84 x^{2} -x^3)=0 \\ \\ \Rightarrow 84 \times 2x - 3 x^{2} =0\\ \\ \Rightarrow 168x-3x^2=0\\ \\ \Rightarrow 3x(56-x)=0\\ \\ \Rightarrow x=0\ or\ 56$

For maximum, $\frac{d^2p}{dx^2} \ \textless \ 0$

$\frac{d^2p}{dx^2}= \frac{d^2(84 x^{2} -x^3)}{dx^2}= \frac{d}{dx}(168x-3x^2)=168-6x \\ \\at\ x=0\\ 168-6x=168-6 \times 0=168\ \textgreater \ 0\ \ \ (condition\ not\ satisfied) \\ \\at\ x=56\\ 168-6x=168-6 \times 56=-168\ \textless \ 0\ \ \ (condition\ satisfied)$

So x = 56,
y = 84 - x = 84- 56 = 28

So the two parts are (28,56)