Question

Maximize Q = xy, where x and y are positive numbers such that x + 3y2 = 25. The maximum value of Q is __ and occurs at x = __ and y = __

(Type exact answers in simplified form.)

Answer 1

To maximize Q = xy under the constraint x + 3y2 = 25, we substitute for x to get Q as a function of y only, then differentiate, set the derivative to zero, and solve for y. Choosing the positive solution y = 5/3, we find x = 50/3. Substituting these values into Q gives the maximum value of Q = 250/9.

Step-by-step explanation:

To maximize the function Q = xy given the constraint x + 3y2 = 25, we can use the method of Lagrange multipliers or substitution. For the high school level, the substitution method is often more accessible. First, solve for one variable in terms of the other using the constraint. Let's solve for x:

x = 25 - 3y2

Now, substitute this expression into our function to maximize Q:

Q = y(25 - 3y2)

Expand and express Q as a function of y only:

Q = 25y - 3y3

To find the maximum, take the derivative with respect to y and set it equal to zero:

Q' = 25 - 9y2

0 = 25 - 9y2

Solve for y:

y = \(\pm\sqrt{\frac{25}{9}}\) = \(\pm\frac{5}{3}\)

Since y must be positive, we discard the negative solution and keep y = \(\frac{5}{3}\). Next, substitute y back into the constraint to solve for x:

x = 25 - 3\(\left(\frac{5}{3}\right)^2\) = 25 - 3(\(\frac{25}{9}\)) = 25 - \(\frac{75}{9}\) = \(\frac{225}{9} - \frac{75}{9}\) = \(\frac{150}{9}\)

x = \(\frac{50}{3}\)

Finally, substitute both x and y into Q to determine the maximum value:

Q = \(\frac{50}{3}\)\(\times\frac{5}{3}\) = \(\frac{250}{9}\)

The maximum value of Q is \(\frac{250}{9}\) and it occurs at x = \(\frac{50}{3}\) and y = \(\frac{5}{3}\).

Answer 2

The maximum value of the function Q = xy, given the constraint x + 3y^2 = 25, is 125/9 and occurs at x = 25/3 and y = 5/3.

Step-by-step explanation:

To maximize the function Q = xy, subject to the constraint x + 3y^2 = 25 with x and y being positive numbers, we can use the method of Lagrange multipliers or substitute for x in the function Q using the constraint. Let's use substitution for simplicity.

From the constraint, we have x = 25 - 3y^2. Substituting this into the function Q gives us Q(y) = y(25 - 3y^2). To find the maximum value, we take the derivative of Q with respect to y and set it equal to 0:

Q'(y) = 25 - 9y^2

Setting the derivative equal to zero gives us:

25 - 9y^2 = 0

9y^2 = 25

y^2 = 25/9

y = 5/3 or y = -5/3 (rejecting the negative value as y is positive).

Substituting y = 5/3 back into the constraint gives us x = 25 - 3(5/3)^2, which simplifies to x = 25 - 3(25/9) or x = 25 - 75/9 = 25 - 25/3 = 50/3 - 25/3 = 25/3.

Therefore, the maximum value of Q occurs at x = 25/3 and y = 5/3, and the maximum value is Q = (25/3)*(5/3) = 125/9.