Question

4. If x, y, and z are nonnegative numbers such that x + y = 5 and x + z = 3,
find the largest possible value of x^3 + (3/2)y^2 + z^3.

Answer 1

\[
\text{Let } S = x^3 + \frac 32y^2 + z^3 \\
x + y = 5; ~~~ y = 5 - x \\
x + z = 3; ~~~ z = 3 - x \\
S = x^3 + \frac 32(5-x)^2 + (3-x)^3 \\
\text{Maximize S.} \\
\]

Answer 2

Thank you! I was doing some trippy stuff on this problem, haha.

Answer 3

But this requires some extra work than a typical maximization problem.
dS/dx = 0 yields x = 2 which happens to be a minimum!

Answer 4

oh noooooooooo

Answer 5

So we have to check endpt 0 since there's no endpt to the right?

Answer 6

"x, y, and z are non-negative numbers" implies:
\(x \ge 0\), \(y \ge 0\), \(z \ge 0\)
x + y = 5; x = 5 - y
Minimum y is 0. So maximum x is 5.
x + z = 3; x = 3 - z
Minimum z is 0. So maximum x is 3.

Therefore, \(0\le x \le 3\).
Check S at x = 0 and x = 3 and see which gives max S.

Answer 7

Oooh okay, thanks, I'll try that right now

Answer 8

S(0) = 64.5 and S(3) = 33. So max at (0, 64.5)!