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AleksandrR [38]
2 years ago
11

David opened a coffee shop and sold 60 mochas the first day at $2 per cup. He wants to increase the price per cup to increase hi

s revenue. He found out that for every $0.25 increase, x, in the price per cup, the number of cups he sold decreased by 2 per day.
How can David find the equation which represents his daily revenue, in dollars, from mocha sales when the price is increased x times?

A. Multiply (60 − 2x) and (2 + 0.25x) to create the equation y = -0.5x2 + 19x + 120.

B. Multiply (60 − 2x) and (2 + 0.25x) to create the equation y = -0.5x2 + 11x + 120.

C. Multiply (60 − 0.25x) and (2 + 2x) to create the equation y = -0.5x2 + 119.5x + 120.

D. Multiply (60 − 0.25x) and (2 + 2x) to create the equation y = -0.5x2 + 120.5x + 120.
Mathematics
2 answers:
viktelen [127]2 years ago
7 0

Answer:

B. Multiply (60 − 2x) and (2 + 0.25x) to create the equation y = -0.5x2 + 11x + 120

Step-by-step explanation:

x is the number of price increases of 0.25 each, so (2 + 0.25x) will be the price after x increases.

2 cups remain unsold for every increase in price, so for x increases, 2x cups remain unsold. Then the number sold is (60 -2x).

Revenue is the product of price and quantity sold, so is ...

... revenue = (60 -2x)(2 +0.25x) . . . . . . matches selections A or B

The product of these binomials is ...

... 60·2 +60·0.25x -2x·2 -2x·0.25x

... = -0.5x² +11x +120 . . . . . . . . . . . . . . . . matches selection B

Alika [10]2 years ago
4 0

He wants to increase the price per cup to increase his revenue. He found out that for every $0.25 increase, x, in the price per cup,

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Answer:

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Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

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Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

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