Which number is rational, an integer, and a real number?

Shakespear's Pizza sells 1 comma 100 large Vegi Pizzas per week for $16 a pizza. When the owner offers a $5 discount, the week

zimovet [89]

Answer:

(a) A(x) = 100x + 1100

(b) x = 2.5 USD

(c) A(x) = 100x + 1100 and x = -1.5 USD

Step-by-step explanation:

(a) Since there's a linear relation between weekly sales A(x) and the discount, we can model it as the following expression

A(x) = mx + b

where x is the discount, m is the slope and b is the intersect of the linear equation. We can solve for slope first

$m = \frac{s_2 - s_1}{x_2 - x_1} = \frac{1600 - 1100}{5 - 0} = \frac{500}{5} = 100$

To solve for b, we can just plug in m = 100, x = 0 and A(0) = 1100

1100 = 100*0 + b

b = 1100

Therefore A(x) = 100x + 1100

With discount x, the actual price would then be p - x. Where p is the original undiscounted pricing ($16 in this case)

And with sale A(x), the revenue would then be

$R = (p - x)A(x) = (p - x)*(100x + 1100)$

$R = 100xp + 1100p - 100x^2 - 1100x$

$R = -100x^2 + (100p - 1100)x + 1100p$

To find the maximum value of this, we can take the 1st derivative and set it to 0

R' = -200x + 100p - 1100 = 0

200x = 100p - 1100

$x = \frac{100p - 1100}{200} = 0.5p - 5.5$

when p = $16 then x = 0.5*16 - 5.5 = 8 - 5.5 = 2.5 USD

(c) when the price p is $8 then A(x) would sill be 100x + 1100 because it doesn't depend on p, but the discount. But to maximize the revenue, x will need to be

0.5p - 5.5 = 0.5*8 - 5.5 = 4 - 5.5 = -1.5 USD

So the Pizza owner would need to raise the price by 1.5 USD because originally the pizza is already inexpensive.

4 0

3 years ago

What is 3.1. Divided by 4

MariettaO [177]

The answer to that is 0.775

4 0

3 years ago

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$