Question

Maximize Q=xy, where x and y are positive numbers such that x+28/3 y²=7. Write the objective function in terms of y. Q= (Type an expression using y as the variable.)

Answer 1

To maximize Q=xy, the objective function in terms of y is Q = 7y - \frac{28}{3}y^3.

Step-by-step explanation:

To maximize Q=xy, we need to write the objective function in terms of y. First, solve the given equation for x in terms of y:

x = 7 - \frac{28}{3}y^2

Now substitute this expression for x into the objective function:

Q = xy = y(7 - \frac{28}{3}y^2) = 7y - \frac{28}{3}y^3

So, the objective function in terms of y is Q = 7y - \frac{28}{3}y^3.

Answer 2

The objective function Q = xy is maximized subject to the constraint x + (28/3)y² = 7 by expressing x in terms of y and then substituting into Q to get Q = 7y - (28/3)y³. We then take the derivative, set it to zero, and solve for y to find the maximum value while ensuring that x and y are positive.

Step-by-step explanation:

To maximize the objective function Q = xy given the constraint x + (28/3)y² = 7, we first need to express Q in terms of y only. Solving the constraint for x gives us:

x = 7 - (28/3)y²

Now, we can substitute this expression for x into the objective function Q:

Q = (7 - (28/3)y²)y

Q = 7y - (28/3)y³

This is our objective function in terms of y. To find the maximum value of Q, we would need to take the derivative of this function with respect to y, set it equal to zero, and solve for y. We must choose the solution that leads to a maximum (and not a minimum or saddle point) and satisfies the original constraint and the condition that x and y are positive numbers.