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3 years ago

Please help me with this word problem!!!

Mathematics

1 answer:

Inessa [10]3 years ago

7 0

Answer:

The number of customers needed to break even is

$\simeq 6775.$

Step-by-step explanation:

Rent per month = $ 7200

Rent per year, $ $7200 \times 12$

= $ 86400 -------------(1)

Insurance per year = $ 4200 ---------(2)

Total direct cost for 2 servers (withoout tax) per year,

= $ $2 \times 15 \times 8 \times 5 \times 50$

= $ 60000

Total cost for helper ( without tax) per year,

= $ $4 \times 15 \times 5 \times \50$

= $ 15000

Total cost for employees per year (without tax),

= $ (60000 +15000) = $ 75000

Total cost per year for employees including tax,

$ $\frac {(75000 \times (100 + 6.2 + 1.45)}{100}$

= $ $\frac {(75000 \times 107.65 }{100}$

= $ 80737.5 -----------------------(3)

So,

total cost for running the restaurant (except food)

= (1) + (2) + (3) = $ $(86400 + 4200 + 80737.5)$

= $ 171337.5

Average total profit per customer,

= ( Average per customer payment - average cost of food per customer)

= $ (45.47 - 20.18)

= $ 25.29

So, total number of customer needed to break even,

= $\frac {171337.5}{25.29}$

\simeq 6775

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25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

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We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

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Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

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According with this theorem

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where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

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