Question

X>0, xy=12Найти наименьшее значение выражения х+3у

Answer 1

Y=12/x   x+3y=x+36/x
y'=1-36/x²   36/x²=1  x²=36   x=6

Answer 2

To find the minimum value of the expression x + 3y given the conditions x > 0 and xy = 12, we can use the method of substitution.

First, let's solve the equation xy = 12 for y in terms of x:

xy = 12 y = 12/x

Now, substitute this value of y into the expression x + 3y:

x + 3(12/x) x + 36/x

To find the minimum value, we can take the derivative of this expression with respect to x and set it equal to zero:

d/dx (x + 36/x) = 1 - 36/x^2 = 0

Solving this equation for x, we get:

1 - 36/x^2 = 0 36/x^2 = 1 x^2 = 36 x = ±6

Since x > 0 according to the given conditions, we take x = 6.

Substituting this value back into the expression x + 3y, we can find y:

x + 3y = 6 + 3y

Now, substitute xy = 12 into this equation:

6 + 3y = 12 3y = 6 y = 2

So, the minimum value of the expression x + 3y is obtained when x = 6 and y = 2. Plugging these values into the expression, we have:

6 + 3(2) = 6 + 6 = 12

Therefore, the minimum value of x + 3y is 12.