# 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 of two positive real numbers x and y is 20. Find...

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 of two positive real numbers x and y is 20. Find the minimum possible value of their sum. Use differentiation with the notation dy/dx.

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### 4 Answers

b) The product of two positive real numbers x and y is 20. Find the minimum possible value of their sum. Use differentiation with the notation dy/dx.

Given: xy = 20. Maximize: x + y

Rewrite the equation we wish to maximize in terms of one variable. x is most convenient:

x + (20/x)

A maximum occurs when the first derivative is zero and the second derivative is greater than 0.

Take the first derivative:

dy/dx = 1 - (20/x^2) = (x^2 - 20) / (x^2)

Set it equal to zero: x^2 - 20 = 0, Therefore x = sqrt(20) or -sqrt(20)

Take the second derivative:

d2y/dx2 = 40/x^3

If we were to plug in sqrt(20) into that second derivative, it would be positive so we know that is the minimum.

The minimum value occurs when x = sqrt(20)

a) Given: x + y = 12. Maximize: xy.

y = 12 - x

Rewrite the product in terms of one variable: x (12 - x) = 12x - x^2

Remember that a max occurs where the first derivative is 0 and the second derivative is less than 0.

Take the first derivative: dy/dx = 12 - 2x

Set to zero: 12 - 2x = 0 Therefore, x = 6

Take the second derivative: d2y/dx2 = -2

-2 is always less than zero.

The product is maximized at x = 6.

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)

x+y = 12. To find the maximum value xy.

Since x+y = 12, y = 12-x

Product P = xy = x(12-x).

By calculus, for the maximum of P(x) is at x= c for which dP/dx = 0 and d2P/dx^2 < 0.

dP/dx = {x(12-x)}' = {12x-x^2}' = 12-2x. Equating to zero we get : 12-2x = Or 2x=12. Or x= 12/2 = 6.

d2P/dx^2 = {12-2x}' = -2. d2p/dx^2 < 0 for all x .

So for x = 6. the product is maximum.

So maximum of the product , x*(12-x) = 6(12-6) = 36 for x= 6.

(ii)

Let 20 be the product of 2 numbers. Then if x is one number, the other number is 20/x.

Let y betheir sum. So y (x)= x+20/x.

For the minimum value of y(x), for x= c, forwhich dy/dx = 0 and d2/dx^2 > 0.

dy/dx = (x+20/x)' = 1-20/x^2. Equating to zero, 1-20/x^2 = 0. Or x^2-20 = 0. x = sqrt20.

d2y/dx^2 = {1-1/x^2}' = 0- (-2)1/x^3 =1/x^3 . Put x= sqrt20. Then {d2y/dx^2 at x=sqrt20} = 2/(sqrt20)^3 > 0.

Therefore y(sqrt20) = sqrt20 +20/sqrt20 = 2sqrt20 = 4sqrt5 is the least sum of two numbers whose product is 20.