Question

Use differentials to approximate the change in cost corresponding to an increase in sales (or production) of one unit. Then compare this with the actual change in cost. (Round your answers to two decimal places.) Function x-Value C = 0.075x2 + 6x + 7 X = 10 dollars dc = AC = dollars Need Help? Raadi Wis This to a Tutor 2. (-/2 Points] DETAILS LARAPCALC10 3.8.016. MY NOTES PRACTICE ANOTHER Use differentials to approximate the change in revenue corresponding to an increase in sales (or production) of one unit. Then compare this with the actual change in revenue Function x-Value R = 50x - 1.5x2 x = 15 dR dollars AR = dollars Need Help? Read Talk to a Tutor

Answer 1

a)

we have

C = 0.075.c? +61 + 7

dC dr d -0.075.c? + 6x + 7 dc

dC = 0.15.1 + 6 dr

dC = (0.15.0 + 6) dc

put x = 10 and dx = 1,

$dC=(0.15(10)+6)(1)$

$dC=7.5$ .......................i)

the actual change in cost,

C = f(0) = 0.075x2 + 6.0 +7

so,

$f(10)=0.075(10)^2+6(10)+7=74.5$

$f(11)=0.075(11)^2+6(11)+7=82.075$

now

$\Delta C=f(x+\Delta x)-f(x)$

put x = 10 and del x = 1,

$\Delta C=f(10+1)-f(10)$

$\Delta C=f(11)-f(10)$

$\Delta C=82.075-74.5$

$\Delta C=7.575$

$\Delta C=7.58$ ...........................ii)

b)

a)

we have

R= 50.1 – 1.5.12

$\frac{dR}{dx}=\frac{d}{dx}50x-1.5x^2$

$\frac{dR}{dx}=50-3x$

$dR=(50-3x)dx$

put x = 15 and dx = 1,

$dR=(50-3(15))(1)$

$dR=5$  .......................i)

the actual change in revenue,

$R=f(x)=50x-1.5x^2$

so,

$f(15)=50(15)-1.5(15)^2=412.5$

$f(16)=50(16)-1.5(16)^2=416$

now

$\Delta R=f(x+\Delta x)-f(x)$

put x = 15 and del x = 1,

$\Delta R=f(15+1)-f(15)$

$\Delta R=f(16)-f(15)$

$\Delta R=416-412.5$

$\Delta R=3.5$ ...........................ii)