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GaryK [48]
3 years ago
5

The revenue function for a production by a theatre group is R(t) = -50t^2 + 300t where t is the ticket price in dollars. The cos

t function for the production is C(t) = 600-50t. Determine the ticket price that will allow the production to break even.
Mathematics
2 answers:
Mkey [24]3 years ago
7 0
Break even is the value of t where revenue=cost or R(t)=C(t)

set equal each other

-50t^2+300t=600-50t
multiply both sides by -1
50t^2-300t=50t-600
divide both sides by 50
t^2-6t=t-12
minus t-12 from both sides
t^2-7t+12=0
factor
(t-4)(t-3)=0
set each to zero
t-4=0
t=4

t-3=0
t=3
the cost is $3 or $4

it will first break even at t=3$
worty [1.4K]3 years ago
5 0
The break even exists when the cost of the process is equal to the revenue of the selling. In this case, -50t^2 + 300t  = 600-50t; <span>50t^2 -350 t  + 600 = 0. t is the number of tickets. In this problem, the breakeven point is at t equal to 4 and 3 tickets. the ticket price should be 600- 50*3 equal to 450 dollars as this is the greater one between the two to gain more revenue after break even</span>
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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0
3 years ago
What percent is the change in the price if: a The price was $100 and now it is $1250?
KatRina [158]

Answer:

The increase is 1150.

11.5 times the original value, or 1150% of it. That's a big increase, and I would not buy that thing, whatever it is.

8 0
3 years ago
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