Find two positive real numbers x and y such that they add to 120 and x^(2)y is as large as possible

Well, I could tell you a joke about numbers, but let's get serious for a moment. To find the numbers x and y that maximize the expression x^2*y, we'll use a bit of calculus.

Let's express x in terms of y, based on the given condition that x + y = 120. We can rewrite it as x = 120 - y.

Now we substitute this value of x into the expression x^2*y:
(120 - y)^2 * y

To find the maximum value of this expression, we can take the derivative with respect to y and set it equal to zero. Let's crunch some numbers in the next step!

Calculating... Crunch, crunch, crunch...

Okay, let me check my calculations again.

Ah, sorry! It looks like there is no maximum value for the expression x^2*y. As y approaches infinity, the expression will also increase without bound, so there is no largest value. So unfortunately, I cannot find a specific pair of numbers that satisfies your conditions.

To find two positive real numbers x and y such that they add to 120 and x^(2)y is as large as possible, we can use the method of maximizing functions.

Let's denote x as one of the numbers and y as the other.

Step 1: Express one variable in terms of the other using the given condition x + y = 120.

Let's use y = 120 - x.

Step 2: Create a function to maximize. In this case, we want to maximize x^(2)y.

Let's substitute the expression for y into the function to obtain f(x) = x^(2)(120 - x).

Step 3: Find the critical points by taking the derivative of the function and setting it equal to zero.

Differentiating f(x) with respect to x:

f'(x) = 2x(120 - x) - x^(2)(-1)
= 240x - 3x^(2)

Setting f'(x) = 0:

240x - 3x^(2) = 0

Step 4: Solve the equation obtained in step 3 to find the critical points.

Factoring out the common factor of x:

x(240 - 3x) = 0

Setting each factor equal to zero and solving for x:

x = 0 or 240 - 3x = 0

For x = 0, y = 120 - x = 120 - 0 = 120.

For 240 - 3x = 0, solving for x:

240 - 3x = 0
3x = 240
x = 80

Substituting x = 80 into y = 120 - x,

y = 120 - 80 = 40.

Step 5: Analyze the critical points to identify any maximum or minimum values.

To determine if the critical points are a maximum or minimum, we need to compare the value of x^(2)y at the critical points with values outside the given range.

When x = 0, x^(2)y = 0.

When x = 80 and y = 40, x^(2)y = 80^(2) * 40 = 256,000.

Since x^(2)y is not defined for negative x values, we only need to compare x^(2)y = 256,000 with x^(2)y at the end of the given range, i.e., x = 120 and y = 0.

When x = 120 and y = 0, x^(2)y = 120^(2) * 0 = 0.

Comparing x^(2)y = 256,000 with x^(2)y = 0, we see that x = 80 and y = 40 give the largest value for x^(2)y within the given range.

Therefore, the two positive real numbers x and y that add to 120 and x^(2)y as large as possible are x = 80 and y = 40.

To find two positive real numbers x and y such that they add to 120 and x^(2)y is as large as possible, we can use calculus to solve this optimization problem.

Let's define the objective function f(x, y) = x^2y. We want to maximize this function subject to the constraint x + y = 120.

To solve this problem using calculus, we need to find the critical points. First, let's express the constraint as y = 120 - x.

Substituting y = 120 - x into the objective function, we have f(x) = x^2 (120 - x).

Differentiating f(x) with respect to x, we get f'(x) = 240x - 3x^2.

Setting f'(x) equal to zero to find critical points:

240x - 3x^2 = 0
3x^2 - 240x = 0
3x(x - 80) = 0

This equation has two solutions: x = 0 and x = 80.

Now, we need to check the endpoints and critical points to determine the maximum value of f(x).

When x = 0, y = 120 - x = 120. So, one solution is (0, 120).

When x = 80, y = 120 - x = 40. So, another solution is (80, 40).

To ensure that this is indeed the maximum, we can compare the values of f(x) at these points:

f(0, 120) = (0^2)(120) = 0
f(80, 40) = (80^2)(40) = 256,000

The maximum value of x^2y is 256,000, which occurs when x = 80 and y = 40.

Therefore, the two positive real numbers that satisfy the given conditions are x = 80 and y = 40.

x+y = 120, so y = 120-x

z = x^2y = x^2(120-x)
dz/dx = 240x - 3x^2 = 3x(80-x)
absolute maximum of x^2y is at x=80

so, x=80, y=40

Note that this is very like the fencing problems, where the length is divided equally among the lengths and widths. Here, the powers of x and y are the key as to how to divide the 120: 40 for x, 40 for x, 40 for y.