Question

DIVIDE THE NUMBER 20 INTO TWO PARTS SUCH THAT THE PRODUCT I S MAXIMUM

Answer 1

20 can be written as ( 10 + 10 ).
and 10 x 10 = 100.
In this case, the product will be maximum.


Others numbers maybe = 
 * 11 + 9 = 20
11 x 9 = 99.

* 18 + 2 = 20
18 x 2 = 36

*17 + 3. = 20
17 x 3 = 51.

* 16 + 4 = 20
16 x 4 = 64

* 15 + 5 = 20
15 x 5 = 75.

* 14 + 6 = 20
14 x 6 = 84.

* 13 + 7 = 20
13 x 7 = 91.

* 12 + 8 = 20
12 x 8 = 96.

* 11 + 9 = 20.
11 x 9 = 99.

* 10 + 10 = 20.
10 x 10 = 100.

Hence by hit and trial method it's proved that ( 10 + 10 ) is the answer.

Answer 2

Let one part = x
other part = (20-x)

their product, p = x(20-x) = 20x - x²
For the product, p to be maximum

$\frac{dp}{dx}=0\\\\ \Rightarrow \frac{d}{dx}(20x- x^{2} )=0\\\\ \Rightarrow 20-2x=0\\\\ \Rightarrow 2x = 20\\\\ \Rightarrow x = \frac{20}{2}\\\\ \Rightarrow x=10\\ \\20-x = 20-10 = 10$

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).