Jawab:
Penjelasan dengan langkah-langkah:
turunan
maks min
y' =0
__
soal 1
Biaya produksi B(x) = x² +100x + 100
Harga penjualan/bks = 220 - 2x
P(x) = x(220 - 2x) = 220x -2x²
Untung =L
L(x) = P(x) - B(x)
L(x) = 220x - 2x² - (x² +100x + 100)
L(x)= 220 x - 2x² - x² - 100x - 100
L(x)= -3x²+ 120 x - 100
L'(x)= 0
-6x +120 =0
x= 20
keuntungan maks
L(20)= - 2(20)² + 120(20) - 100
L(20) = -800 + 2400 - 100
L(20) = 1.500 (dlm ribuan)
keuntungan maks =Rp 1.500.000
soal 2
Keliling = 2x + 24
2(p + l) = 2 (x + 12)
p + l = x + 12
p = x + 12 - l
p(x) = x + 12 - (8 - x)
p(x)= x + 12 - 8 + x
p(x)= 2x + 4
Luas = p l
L(x) = p(x) . l(x)
L(x)= (2x + 4)(8 - x)
L(x) = 16x - 2x² + 32 - 4x
L(x)= -2x² + 12 x + 32
L'(x) =0
-4x + 12= 0
x= 3
luas maks ,jika p= 2x+ 4
p = 2(3) + 4
p = 10 cm
Jawaban:
1. Biaya = (x²+100x+100) ribu rupiah = 2x+100
Harga jual = (220-2x) ribu rupiah = (220-2x)x=220x-2x² = -4x+220
Keuntungan maksimum=...
Keuntungan =
Pendapatan - Biaya
(-4x+220)-(2x+100)=0
-4x+220-2x-100=0
-6x+120=0
-6x=-120
x=-120/-6
x=20
Pendapatan = 220x-2x² = 220(20)-2(20)²=4400-2(400)=4400-800= 3600
Biaya = x²+100x+100=
20²+100(20)+100=400+2000+100=2500
Keuntungan = 3600 - 2500 = 1.100 ribu rupiah
atau Rp 1.100.000,₋
2. Persegi Panjang
Keliling = (2x+24) meter
Lebar = (8-x) meter
Panjang taman agar luasnya maksimum=...
Keliling =
2(P+L) = (2x+24)
2(P+(8-x) = (2x+24)
(P+8-x) = (2x+24)/2
(P+8-x) = x+12
(P) = x+x+12-8
(P) = 2x+4
Luas = P x L
(2x+4)(-x+8)
(-2x²-4x+16x+32)=0
(-2x²+12x+32)=0
Turunan
-4x+12=0
-4x=-12
x=-12/-4
Panjang = 2x+4 = 2(3)+4=6+4=10 meter
Lebar = 8-x = 8 - 3 = 5 meter
Pembuktian
Keliling = 2x+24 = 2(3)+24 = 6 + 24 = 30 meter
2(P+L) = 2(10+5) = 2(15) = 30 meter
3. Panjang kawat = 100 cm
Keliling = Persegi panjang + 1/4 lingkaran
2(P+L) + 1/4(2xπxr) = 100
2(P+L) + 1/4(2x3,14xL) = 100
2(P+L) + 1/4(6,28L) = 100
2(P+L) + 1,57L = 100
2P + 2L + 1,57L = 100
2P + 3,57L = 100
2P = -3,57L + 100
P = (-3,57L + 100)/2
P = -1,785L + 50
Luas = Persegi panjang + 1/4 lingkaran
(P x L ) + 1/4(πxrxr)
(-1,785L+ 50)L + 1/4(3,14xLxL)
(-1,785L²+50L + 1/4(3,14xL²)
(-1,785L²+50L + 0,785L²)
(-1L²+50L)
(-L²+50L) = 0
(-2L+50)=0
(-2L)=-50
(L) =-50/-2
(L)=25
Luas daerah maksimum adalah = -(25)²+50(25) = -625+1250 = 625 cm²
Catatan
Panjang = -1,785 (25) + 50 = -44,625 + 50 =5,375 cm
Lebar = 25 cm
2P+3,57L=100
2(5,375) + 3,57(25) = 100
10,75 + 89,25 = 100
100 = 100 (sama)
4. Panjang = 24 cm
Panjang = 24-x-x=24-2x
Lebar = 9 cm
Lebar = 9-x-x=9-2x
Disudut karton dipotong persegi x cm
Kotak tanpa tutup
Volume maksimum kotak adalah...
Volume
P x L x T
(24-2x)(9-2x)x
(-2x+24)(-2x+9)x
(4x²-48x-18x+216)x
(4x²-66x+216)x
(4x³-66x²+216x)
(12x²-132x+216)=0
(x²-11x+18)=0
(x-9)(x-2)=0
(x₁)=9 atau (x₂)=2 (√ )
p=24-2x
=24-2(9)
=24-18
=6
atau
=24-2(2)
=24-4
=20
l=9-2x
=9-2(9)
=9-18
=-9 (tidak)
=9-2(2)
=9-4
=5
p x l x t
20 x 5 x 2
20 x 10
200 cm³
Demikian
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Jawab:
Penjelasan dengan langkah-langkah:
turunan
maks min
y' =0
__
soal 1
Biaya produksi B(x) = x² +100x + 100
Harga penjualan/bks = 220 - 2x
P(x) = x(220 - 2x) = 220x -2x²
Untung =L
L(x) = P(x) - B(x)
L(x) = 220x - 2x² - (x² +100x + 100)
L(x)= 220 x - 2x² - x² - 100x - 100
L(x)= -3x²+ 120 x - 100
L'(x)= 0
-6x +120 =0
x= 20
keuntungan maks
L(20)= - 2(20)² + 120(20) - 100
L(20) = -800 + 2400 - 100
L(20) = 1.500 (dlm ribuan)
keuntungan maks =Rp 1.500.000
soal 2
Keliling = 2x + 24
2(p + l) = 2 (x + 12)
p + l = x + 12
p = x + 12 - l
p(x) = x + 12 - (8 - x)
p(x)= x + 12 - 8 + x
p(x)= 2x + 4
Luas = p l
L(x) = p(x) . l(x)
L(x)= (2x + 4)(8 - x)
L(x) = 16x - 2x² + 32 - 4x
L(x)= -2x² + 12 x + 32
L'(x) =0
-4x + 12= 0
x= 3
luas maks ,jika p= 2x+ 4
p = 2(3) + 4
p = 10 cm
Jawaban:
Turunan
Penjelasan dengan langkah-langkah:
1. Biaya = (x²+100x+100) ribu rupiah = 2x+100
Harga jual = (220-2x) ribu rupiah = (220-2x)x=220x-2x² = -4x+220
Keuntungan maksimum=...
Keuntungan =
Pendapatan - Biaya
(-4x+220)-(2x+100)=0
-4x+220-2x-100=0
-6x+120=0
-6x=-120
x=-120/-6
x=20
Pendapatan = 220x-2x² = 220(20)-2(20)²=4400-2(400)=4400-800= 3600
Biaya = x²+100x+100=
20²+100(20)+100=400+2000+100=2500
Keuntungan = 3600 - 2500 = 1.100 ribu rupiah
atau Rp 1.100.000,₋
2. Persegi Panjang
Keliling = (2x+24) meter
Lebar = (8-x) meter
Panjang taman agar luasnya maksimum=...
Keliling =
2(P+L) = (2x+24)
2(P+(8-x) = (2x+24)
(P+8-x) = (2x+24)/2
(P+8-x) = x+12
(P) = x+x+12-8
(P) = 2x+4
Luas = P x L
(2x+4)(-x+8)
(-2x²-4x+16x+32)=0
(-2x²+12x+32)=0
Turunan
-4x+12=0
-4x=-12
x=-12/-4
x= 3
Panjang = 2x+4 = 2(3)+4=6+4=10 meter
Lebar = 8-x = 8 - 3 = 5 meter
Pembuktian
Keliling = 2x+24 = 2(3)+24 = 6 + 24 = 30 meter
2(P+L) = 2(10+5) = 2(15) = 30 meter
3. Panjang kawat = 100 cm
Keliling = Persegi panjang + 1/4 lingkaran
2(P+L) + 1/4(2xπxr) = 100
2(P+L) + 1/4(2x3,14xL) = 100
2(P+L) + 1/4(6,28L) = 100
2(P+L) + 1,57L = 100
2P + 2L + 1,57L = 100
2P + 3,57L = 100
2P = -3,57L + 100
P = (-3,57L + 100)/2
P = -1,785L + 50
Luas = Persegi panjang + 1/4 lingkaran
(P x L ) + 1/4(πxrxr)
(-1,785L+ 50)L + 1/4(3,14xLxL)
(-1,785L²+50L + 1/4(3,14xL²)
(-1,785L²+50L + 0,785L²)
(-1L²+50L)
Turunan
(-L²+50L) = 0
(-2L+50)=0
(-2L)=-50
(L) =-50/-2
(L)=25
Luas daerah maksimum adalah = -(25)²+50(25) = -625+1250 = 625 cm²
Catatan
Panjang = -1,785 (25) + 50 = -44,625 + 50 =5,375 cm
Lebar = 25 cm
Pembuktian
2P+3,57L=100
2(5,375) + 3,57(25) = 100
10,75 + 89,25 = 100
100 = 100 (sama)
4. Panjang = 24 cm
Panjang = 24-x-x=24-2x
Lebar = 9 cm
Lebar = 9-x-x=9-2x
Disudut karton dipotong persegi x cm
Kotak tanpa tutup
Volume maksimum kotak adalah...
Volume
P x L x T
(24-2x)(9-2x)x
(-2x+24)(-2x+9)x
(4x²-48x-18x+216)x
(4x²-66x+216)x
(4x³-66x²+216x)
(12x²-132x+216)=0
(x²-11x+18)=0
(x-9)(x-2)=0
(x₁)=9 atau (x₂)=2 (√ )
p=24-2x
=24-2(9)
=24-18
=6
atau
p=24-2x
=24-2(2)
=24-4
=20
l=9-2x
=9-2(9)
=9-18
=-9 (tidak)
atau
l=9-2x
=9-2(2)
=9-4
=5
Volume
p x l x t
20 x 5 x 2
20 x 10
200 cm³
Demikian
Semoga membantu dan bermanfaat!