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by sweetysiri92

2015-03-27T19:14:50+05:30

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price = p
demand = x
we need to write demand as a function of price. So f(p) = x.

1.
p = 42 - 0.4x ; 0 ≤ x ≤ 105
for x = 0, p = 42
for x = 105, p = 42 - 0.4×105 = 42 - 42 = 0

p = 42 - 0.4x
⇒ p - 42 = -0.4x
⇒ x = -(p-42)/0.4
⇒ x = 105 - 2.5p
⇒ f(p) = 105 - 2.5p ; 0 ≤ p ≤ 105.

2.
p = 125 - 0.02x ; 0 ≤ x ≤ 6250
for x = 0, p = 125
for x = 6250, p = 125 - 0.02×6250 = 125 - 125 = 0

p =125 - 0.02x
⇒ p - 125 = -0.02x
⇒ -x = (p - 125)/0.02
⇒ -x = 50p - 6250
⇒ x = 6250 - 50p
⇒ f(p) = 6250 - 50p ; 0 ≤ p ≤ 125.

3.
p = 50 - 0.5x² ; 0 ≤ x ≤10
for x = 0, p = 50
for x = 10, p = 50 - 0.5×10² = 50 - 50 = 0

p =50 - 0.5x²
⇒ p - 50 = -0.5x²
⇒ 0.5x² = 50 - p
⇒ x² = (50 - p)/0.5 = 100 - 2p
⇒ x = √(100 - 2p)
⇒ f(p) = √(100 - 2p); 0 ≤ p ≤ 50.

4.
p = 180 - 0.8x² ; 0 ≤ x ≤15
for x = 0, p = 180
for x = 15, p = 180 - 0.8×15² = 180-180 = 0

p =180 - 0.8x²
⇒ p - 180 = -0.8x²
⇒ 0.8x² = 180 - p
⇒ x² = (180 - p)/0.8 = 225 - 1.25p
⇒ x = √( 225 - 1.25p)
⇒ f(p) = √( 225 - 1.25p); 0 ≤ p ≤ 180.

5.
p = 25e^{-x/20} ; 0≤x≤20
for x = 0, p = 25
for x = 20, p =  25e^{-20/20} = 25/e

p = 25e^{-x/20}
⇒ p = 25/e^{x/20}
⇒ e^{x/20} = 25/p
⇒ log( e^{x/20} ) = log (25/p)
⇒ x/20 =  log (25/p)
⇒ x = 20 log (25/p)
⇒ f(p) = 20 log (25/p); 0 ≤ p ≤ 25/e

6.
p = 45 - e^{x/4} ; 0≤x≤12
for x = 0, p = 45-1 = 44
for x = 12, p =  45 - e^{12/4} = 45- e³

p = 45 - e^{x/4}
⇒ p - 15= e^{x/4}
⇒ e^{x/4} = 45 - p
⇒ log( e^{x/4} ) = log (45 - p)
⇒ x/4 =  log (45 - p)
⇒ x = 4 log (45 - p)
⇒ f(p) = 20 log (25/p); 45-e³ ≤ p ≤ 44

7.
p =80 - 10 ln x; 1 ≤ x ≤ 30
for x = 1, p = 80 - 0 = 80
for x = 30, p =  80 - 10 ln (30) = 10 (8 - ln 30)

p = 80 - 10 ln x
⇒ p - 80 = - 10 ln x
⇒ 10 ln x = 80-p
⇒ ln x = (80-p)/10 = 8 - 0.1p
⇒ x = e^{8 - 0.1p}
⇒ f(p) = e^{8 - 0.1p}; 10(8- ln 30) ≤ p ≤ 80

8.
p =ln(500 - 5x); 0 ≤ x ≤ 90
for x = 0, p =ln 500
for x = 90, p =  ln(500 - 5×90) = ln 50

p = ln(500 - 5x)
⇒ ln(500 - 5x) = p
⇒ 500 - 5x = e^p
⇒ -5x = e^p - 500
⇒ x = (e^p - 500)/(-5) = 100 - 0.2e^p
⇒ f(p) = 100 - 0.2e^p; ln 50 ≤ p ≤ ln 500