Find two positve real numbers whose product is a maximum and whose sum of the first number and four times the second number is 120.

Question

Find two positve real numbers whose product is a maximum and whose sum of the first number and four times the second number is 120.

This is as far as I've gotten in solving it.

*first number is x.
*second number is y.
*xy=maximum, which is unknown.
*x+4y=120. Solve for x, which comes out to be x=-4y+120.
*Substitute -4y+120 for x in the equation xy, then solve for y. So (-4y+120)(y) = -4y(y-30) = y=0 and y=30.
Please show me what to do from here.

Answer 1

From your calculations, you have found that the possible values for y are 0 and 30. To determine the corresponding values of x, substitute these values back into the equation x = -4y + 120.

1. For y = 0:
x = -4 * 0 + 120
x = 120

So when y = 0, x = 120.

2. For y = 30:
x = -4 * 30 + 120
x = -120 + 120
x = 0

So when y = 30, x = 0.

Therefore, the two positive real numbers that satisfy the given conditions are:
1. x = 0 and y = 30
2. x = 120 and y = 0

Answer 2

To find the two positive real numbers that satisfy both conditions, we need to find the maximum value of their product while satisfying the second equation.

We already obtained one solution, y = 30. Now let's substitute this value back into the second equation to find x:

x + 4y = 120
x + 4(30) = 120
x + 120 = 120
x = 0

So, one pair of positive real numbers that satisfy both conditions is (0, 30). However, note that x = 0 is not positive, so this is not a valid solution. We need to find another pair.

To maximize the product, we can substitute the expression for x in terms of y from the second equation into the equation for the product:

xy = maximum
(-4y + 120)(y) = maximum

To find the maximum value, we can plot a graph, differentiate the equation, and find the critical points. Let's differentiate the equation with respect to y:

d(xy)/dy = d((-4y + 120)(y))/dy
= (-4)(y) + (-4y + 120)(1)
= -4y - 4y + 120
= -8y + 120

To find the critical point, we set the derivative equal to zero:

-8y + 120 = 0
-8y = -120
y = 15

Now, substitute this value of y back into the second equation to find the corresponding value of x:

x + 4y = 120
x + 4(15) = 120
x + 60 = 120
x = 60

Therefore, the second pair of positive real numbers that satisfy both conditions is (60, 15).

In summary, the two positive real numbers whose product is maximized and whose sum of the first number and four times the second number is 120 are (0, 30) and (60, 15).

Answer 3

You know that 4(30)=120, so that would mean x would have to be 0, since x+4y=120. However, the problem states that both x and y must be real numbers. Since 2(2)=4, you need to find a number in which 2(2)(30)=120. What number times 2 will give 30? 15. x+4(15)=120. Solve for x. x=60. Thus, x+4y=120 is 60+4y=120, solving for y, you get y=15. So the first number is 60 and the second number is 15.