Question

Find two numbers x and y such that x y = 96 and xy2 is maximized.

Answer 1

To maximize xy^2 given xy = 96, we can use the AM-GM inequality to find that x = y = sqrt(96).

Step-by-step explanation:

To find two numbers x and y such that xy = 96 and xy^2 is maximized, we can use the AM-GM inequality. Let's represent x and y as the geometric mean and arithmetic mean of two numbers a and b. So, x = sqrt(ab) and y = (a+b)/2. We have xy = sqrt(ab) * (a+b)/2 = (a+b)/2 * sqrt(ab) = (ab + 2sqrt(ab))/2 = (ab + 2(sqrt(a)sqrt(b)))/2. Since ab is a constant value of 96, we want to maximize the value of 2(sqrt(a)sqrt(b)). In this case, we can see that the maximum value occurs when sqrt(a) = sqrt(b), which means a = b. So, we can let a = b = sqrt(96). Therefore, x = y = sqrt(96).

Answer 2

To find two numbers x and y such that their product is 96 and xy^2 is maximized, we set up an optimization problem. By expressing y in terms of x, we obtain an objective function to maximize. Using calculus, we discover that x and y must be equal and both are the square root of 96.

Step-by-step explanation:

The question is asking to find two numbers x and y such that the product of x and y is 96, and the value of x multiplied by y squared is maximized. This is an optimization problem that can be tackled using the method of taking derivatives or using algebraic manipulation since it involves a quadratic expression.

Steps to Maximize xy2

Let's start with the given condition xy = 96.

We need to express y in terms of x, so y = 96/x.

Now, our objective function to maximize is xy2 = x(96/x)2 = 962x-1.

We can find the derivative with respect to x and set to zero to find the critical point.

Solve the derivative equation for x to find the value that maximizes the objective function.

Using calculus or algebraic methods, you can find that the values of x and y that maximize xy2 under the given condition are equal, thus x = y. Plugging x into xy = 96 gives x = y = sqrt(96), so the two numbers are sqrt(96) and sqrt(96).