The sum of two real numbers x and y is 12. Find the maximum value of the product xy. Use differentiation with the notation dy/dx.b) The product...

a)

Given two numbers x and y
such that:

x + y = 12   ...   
(1)

we have to find value of y for which xy is
maximum:

In equation (1) transferring x from left hand side
to tight hand side:

y = 12 - x   ...  
(2)

Using this value of y, we represent xy
as

xy = f(x)= x(12 -
x)

==> f(x) = 12x -
x^2

Differentiating the above
function:

f'(x) = 12 -
2x

Maximum value of f(x) occurs at point for which f'(x) =
0.

Equating f'(x) to 0 we
get:

12 - 2x = 0

==> -
2x = - 12

==> x = -12/12 =
6

Substituting this value of x in equation
(2):

y = 12 - 6 = 6

Therefore,
value of xy is maximum when:

x = 6 and y =
6

The maximum value of xy = 6*6 =
36

b)

Given two numbers x and
y such that:

xy = 20  ...   
(1)

we have to find value of y for which (x + y) is
maximum:

In equation (1) dividing both sides by
x:

y = 20/x   ...   (2)

Using
this value of y, we represent (x + y) as

(x + y) = f(x)= x
+ 20/x

Differentiating the above
function:

f'(x) = 1 -
20/(x^2)

Maximum value of f(x) occurs at point for which
f'(x) = 0.

Equating f'(x) to 0 we
get:

1 - 20/(x^2)

==>
20/(x^2) = 1

Multiplying both sides by
x^2

20 = x^2

x =
20^(1/2)

Substituting this value of x in equation
(2):

y = 20/20(^1/2) =
20(^1/2)

Therefore, value of x + y is minimum
when:

x = 20(^1/2) and y =
20(^1/2)

The minimum value
of

x + y = 20(^1/2) +
20(^1/2)

= 2*20^(1/2) = 4*5^(1/2)