A theater manager graphed weekly profits as a function of the number of patrons and found that the relationship was linear. One week the profit was $10,229 when 1374 patrons attended. Another week 1595 patrons produced a profit of $12,107.50

A. Find a formula for weekly profit, y, as a function of the number of patrons, x.

B. What is the break-even point (the number of patrons for which there is zero profit)? Round your answer to the nearest integer.

C.Find a formula for the number of patrons as a function of profit.

D.If the weekly profit was $15,558.50, how many patrons attended the theater?

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ColinJacobus ColinJacobus · Answer 1 · 2018-09-20T13:05:36+02:00

Given that profit is a linear function of number patrons.

Let p be the profit and x be the number of patrons.

Let the linear relation be p=mx+b ( equation of straight line having slope m)

A)Given profit p=10229$, for x=1374

That is 10229 =m(1374)+b

For x= 1595, profit p=12,107.50

That is 12107.5 = m(1595)+b

Subtracting first equation from second equation

12107.5-10229 = m*1595+b -(m*1374+b)

1878.5 = m*(1595-1374)

221m = 1878.5

[tex]m=\frac{1878.5}{221} = 8.5[/tex]

To find b, we will plugin m=8.5 in first equation.

10229 = 8.5*1374+b

b= 10229-8.5*1374

b=10229-11679= -1450

Hence equation is profit = 8.5x-1450

B) Let us equate profit to 0 and solve for x.

8.5x-1450 = 0

8.5x = 1450

[tex]x=\frac{1450}{8.5} = 170.588[/tex] ≈ 171

Hence break even point is (171,0)

C) p= 8.5x-1450

p+1450 = 8.5x

[tex]x=\frac{p+1450}{8.5}[/tex]

D)Given weekly profit = 15558.5

So, number of patrons x = [tex]\frac{15558.5+1450}{8.5} = 2001[/tex]

≈2001 patrons