Question

Diketahui dua bilangan real positif x dan y. Jika x+2y = 20, maka nilai maksimum dari x^2y adalah
A. 16000/9
B. 16000/27
C. 4000/27
D. 1600/27
E. 400/9

Answer 1

Verified answer

X + 2y = 20 => x = 20 - 2y

x^2.y = (20 - 2y)^2.y
= (400 - 80y + 4y^2).y
= 400y - 80y^2 + 4y^3
Kita cari titik stasioner nya dengan cara diturunkan
400 - 160y + 12y^2 = 0 ====> bagi 4
3y^2 - 40y + 100 = 0
(3y - 10)(y - 10) = 0
y = 10/3 atau y = 10

x = 20 - 2y
y = 10 => x = 20 - 2(10) = 0
Nilai x^2.y = (0)^2 . 10 = 0

y = 10/3 => x = 20 - 2(10/3) = 60/3 - 20/3 = 40/3
Nilai x^2.y = (40/3)^2 . 10/3 = 1600/9 . 10/3 = 16.000/27

Jadi nilai maksimum nya 16.000/27

Answer 2

Verified answer

X + 2y = 20
2y = 20 - x
y = (20 - x )/2

x^2 y
= x^2 . (20- x)/2
= (20x^2 - x^3) / 2

f(x) = (20x^2 - x^3)/2

u = 20x^2 - x^3
u' = 40x - 3x^2
v = 2
v' = 0

f'(x) = (u'v - uv') / (v^2)
= (40x - 3x^2)2 - (20x^2 - x^3)0 / 2^2
= (40x - 3x^2)2 / 4
= (20x - (3/2)x^2)

nilai maksimum dicapai saat f'(x) = 0
(20x - (3/2)x^2) = 0
Kali 2 kedua ruas
40x -3x^2 = 0
Kali - 1 kedua ruas
3x^2 - 40x = 0
x(3x - 40) = 0
x = 0
Atau x = 40/3
Ambil 40/3 supaya menjadi maksimum.

x = 40/3

y = (20 - (40/3)) /2
y = (20)/2 - (40/3)/2
y = 10 - 40/6
y = 20/6

x^2 y
= (40/3)^2 . 20/6
=( 1600 / 9) . (20/6)
= 32000 / 54
= 16000/27

B