Question

Divide the number 20 in to two parts such
that their product is maximum.

this question is of 12th com application of derivatives ​

Answer 1

Answer:

Let one part = x

other part = (20-x)

their product, p = x(20-x) = 20x - x²

For the product, p to be maximum

So, product is maximum when we divide 20 into 2 parts as (10,10).

Note: If you want to show that dividing 20 into (10,10) yields maximum product, not minimum; find the second derivative of p. If the value of second derivative at x=10 comes as negative, then it is maximum. (It comes -2, so it is maximum).

Answer 2

Answer:

Let one part = x

other part = (20-x)

their product, p = x(20-x) = 20x - x²

For the product, p to be maximum

$\begin{gathered}\frac{dp}{dx}=0\\\\ \Rightarrow \frac{d}{dx}(20x- x^{2} )=0\\\\ \Rightarrow 20-2x=0\\\\ \Rightarrow 2x = 20\\\\ \Rightarrow x = \frac{20}{2}\\\\ \Rightarrow x=10\\ \\20-x = 20-10 = 10\end{gathered} </p><p>dx</p><p>dp$

So, product is maximum when we divide 20 into 2 parts as (10,10).

Note: If you want to show that dividing 20 into (10,10) yields maximum product, not minimum; find the second derivative of p. If the value of second derivative at x=10 comes as negative, then it is maximum. (It comes -2, so it is maximum).