Maximize B=XY^2 , where x and y are positive numbers such that X+Y^2=5.

Without using any calculus:

If x + y² = 5, then y² = 5 - x, so that

b = x y² = x (5 - x) = 5 x - x²

Complete the square to get

b = 25/4 - (x - 5/2)²

Then b gets a maximum value of 25/4 when x = 5/2.

Simple calculus method:

With b = 5 x - x², differentiate b with respect to x and set the derivative equal to 0 to solve for critical points:

db/dx = 5 - 2 x = 0   ==>   x = 5/2

Take the second derivative to determine whether b is concave or convex:

d²b/dx² = -2

The second derivative is negative, which means b is concave with respect to x, and this indicates x = 5/2 is the site of a local maximum, which is b (5/2) = 25/4.

More complicated calculus method:

Lagrange multipliers. The Lagrangian is

L (x, y, λ) = x y² - λ (x + y² - 5)

where we assume λ ≠ 0.

Take its derivatives:

dL/dx = y² - λ

dL/dy = 2 x y - 2 λ y

dL/dλ = x + y² - 5

Set each derivative equal to 0 to find the critical points. The first equation tells us y² = λ, and the second equation tells us x = λ, so

x + y² - 5 = 2 λ - 5 = 0

Solving this gives λ = x = y² = 5/2, so that b has an extreme value of

b = (5/2) (5/2) = 25/4

Deciding whether this is a maximum or minimum comes down to computing the Hessian determinant at the critical value. The Hessian is

with determinant |H (x, y)| = -4 y². At the critical point, the determinant is -10 < 0, which indicates a maximum value of 25/4.

2 x t divided by n

step-by-step explanation:

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.