tolong pakai cara.........

nilai maksimum

y = x+ √(p- 2x)

(1) untuk y =4
x + √(p-2x) = 4
√(p-2x) = 4 -x
p - 2x = (4 - x)²
p - 2x = x² -8x + 16
p= x² - 6x + 16

(2) nilai maks ,jika y'  = 0
y = x + √(p-2x)
y = x + (p-2x)¹/²
y' = 0
1  + (1/2)(-2)  (p -2x)¹/²  = 0
1 - (p-2x)¹/² = 0
p - 2x =1
p= 1 + 2x ....(2)
.
(1) = (2)
x² - 6x + 16 = 1 + 2x
x²-8x + 15 = 0
(x - 5)(x - 3)= 0
x = 5  atau x = 3
.
p = 1+2x
untuk x = 5 --> p = 11
unyuk x = 3 --> p = 7