Let the number of reserved tickets = x
Let the number of lawn seats = y
Constraint functions:
Maximum capacity means 
For concert to be held 
means 
Objective functions :
Maximum profit equation p = 65x +40y
Intersection points :
(10000,10000) (20000,0)(2500,2500)(5000,0)
p at (10000,10000) = 65(10000) + 40(10000) = $1050000
p at (20000,0) = 65(20000) + 40(0) = $1300000
p at (2500,2500) = 65(2500) + 40(2500) = $262500
p at (5000,0) = 65(5000) + 40(0) = $325000
Hence maximum profit occurs when all 20000 reserved seats are sold and the profit is $1300000
Please find attached the graph of it.
log(4)x + log(4)(x + 5) = 6
log(4)x + log(4)x + log(20) = 6
log(8)x + log(20) = 6
<u> - log(20) - log(20)</u>
<u>log(8)x</u> = <u>6 - log(20)</u>
log(8) log(8)
x ≈ 0.999
Answer:
Statement 3
Step-by-step explanation:
y = -2/3x-24
Make x the subject
y + 24 = -(⅔)x
x = -(3/2)(y + 24)
Intersection variables
f^-1(x) = -(3/2)(x + 24)
Slope: -3/2
No domain restrictions
y-intercept: x = 0
-(3/2)(0+24)
-36
(0,-36)
x-intercepts: y = 0
0 = -(3/2)(x + 24)
x = -24
(-24,0)
Linear function, range is all real values of y
360 divided by 115 = 3.13. Then you divide the 103 by the 3.13 = 32.90
Answer:
Step-by-step explanation:
its A)
