Find two positive real numbers whose product is a maximum. (enter your answers as a comma-separated list.) the sum of the first and three times the second is 54.

the sum of the first and three times the second is 54.

x + 3y = 54

3y = 54 - x

y= 54 /3 - x / 3 = 18 -x /3

xy = x (18 -x /3) = 18x - x^2 / 3 = - 1/3 x ^ 2 + 18 x

since a<0, it is concave downward, and the vertex is the maximum value. That vertex occurs at x= - b /2a = (-18) / (-2/3) = 12

The first number is x=12.
The second is y = 18 -x /3 = 18- 12/ 3 = 18 - 4 = 14

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oladayoademosu oladayoademosu · Answer 2 · 2021-09-30T13:39:00+02:00

The two positive real numbers are 9, 33

Let x and y be the two positive real numbers.

Their product is f(x,y) = xy and since the sum of the first and three times the second is 54, we have x + 3y = 54

So, x = 54 - 3y

Substituting x into f, we have

f(x,y) = xy

f(y) = (54 - 3y)y

f(y) = 54y - 3y²

f is maximum when df/dy = 0

So, df/dy = d(54y - 3y²)/dy

df/dy = 54 - 6y = 0

54 = 6y

y = 54/6

y = 9

To determine if f is maximum, we differentiate df/dy again.

So, d(df/dy)/dy = d²f/dy²

= d(54 - 6y)/dy

= -6

Since d²f/dy² = -6 < 0, f is maximum at y = 9

Also, since x = 54 - 3y, substituting the value of y into the equation, we have

x = 54 - 3(9)

x = 54 - 21

x = 33

So, the two positive real numbers are 9, 33

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