R(x) = 60x - 0.2x^2
The revenue is maximum when the derivative of R(x) = 0.
dR(x)/dx = 60 - 0.4x = 0
0.4x = 60
x = 60/0.4 = 150
Therefore, maximum revenue is 60(150) - 0.2(150)^2 = 9000 - 4500 = $4,500
Maximum revenue is $4,500 and the number of units is 150 units
The evaluation of the expression, if m = − 4 , n = 1 , p = 2 , q = − 6 , r = 5 , and t = − 2 | 16 + 4 ( 3 q + p ) is 46.
<h3>How can the expression be simplified?</h3>
the given expression is t = − 2 | 16 + 4 ( 3 q + p )
Then since we are given m = − 4 , n = 1 , p = 2 , q = − 6 , r = 5
Then, we can substitute all the given values of the terms into the given expression as :
t = − 2 | 16 + 4 ( 3 q + p )
t= -2 I 16+4(-18+2)
t= -2 I 16+4(-16)
t= -2 I 16 -64
t= -2 I -48
t =46
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Answer:
Five hundred forty thousandths
Answer:
ixgj zgjfjzufztixititdtidtidtdiustut
X = 3/2
Y = -3
-6x-5(3-4x)=6
Y=3-4(3/2)
