K = (2x + 24) meter
lebar = (8 - x) meter ❗
jika L maks, p = ?
K = 2 (panjang + lebar)
(2x + 24) = 2 (p + 8 - x)
½ (2x + 24) = (p + 8 - x)
x + 12 = p + 8 - x
p = x + 12 - 8 + x
p = 2x + 4
Luas = panjang x lebar
Luas = (2x + 4) (8 - x)
L = 16x -2x² +32 -4x
L = -2x² +12x +32
L maks jika L' = 0
L' = -4x + 12 = 0
-4x = -12
x = -12 / -4
x = 3
p = 2(3) + 4
jadi, panjang taman tsb 10 meter.
_____________
Kelas 11
Matematika
Diferensial
Luas maksimal
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diketahui:
K = (2x + 24) meter
lebar = (8 - x) meter ❗
ditanya:
jika L maks, p = ?
jawab:
K = 2 (panjang + lebar)
(2x + 24) = 2 (p + 8 - x)
½ (2x + 24) = (p + 8 - x)
x + 12 = p + 8 - x
p = x + 12 - 8 + x
p = 2x + 4
Luas = panjang x lebar
Luas = (2x + 4) (8 - x)
L = 16x -2x² +32 -4x
L = -2x² +12x +32
L maks jika L' = 0
L' = -4x + 12 = 0
-4x = -12
x = -12 / -4
x = 3
p = 2x + 4
p = 2(3) + 4
p = 10 meter ✔️
jadi, panjang taman tsb 10 meter.
_____________
Kelas 11
Matematika
Diferensial
Luas maksimal