Question

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

Hint:

To find the maximum or minimum of a function, find the first derivative and set it to zero to find critical points. Then use the second derivative test to determine whether each point is a maximum (\(f''<0\)) or a minimum (\(f''>0\)).

Answer 1

Let one part be \(x\) then the other part will be \(84-x\).
Let the product be \( P(x) \). We can define the function as \( P(x) = x^2(84-x) \). \[ f(x) = \mathbf{84x^2 - x^3} \] \[ \therefore f'(x) = 168x - 3x^2 \] For extreme values \( f'(x)=0 \). \[ 168x - 3x^2 = 0 \] \[ \therefore 3x(56-x) = 0 \] \[ \therefore x = \mathbf{0} \text{ OR } x = \mathbf{56} \] Now, \( f''(x) = 168 - 6x \). If \( x=0, f''(0) = 168 - 6(0) = 168>0 \).
\(\therefore\) function attains minimum at \(x=0\).
If \( x=56, f''(56) = \mathbf{168 - 336 = -168}<0 \).
\(\therefore\) function attains maximum at \(x=56\).
Two parts of 84 are \(\mathbf{56}\) and \(\mathbf{28}\).