Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. assume that revenue, upper r left parenthesis x right parenthesisr(x), and cost, upper c left parenthesis x right parenthesisc(x), of producing x units are in dollars. upper r left parenthesis x right parenthesisr(x)equals=4 x4x, upper c left parenthesis x right parenthesisc(x)equals=0.01 x squared plus 0.3 x plus 40.01x2+0.3x+4

185 units

Step-by-step explanation:

Given,

The revenue function is,

Cost function,

,

Where,

x = number of units produced.

Thus, profit = revenue - cost

Differentiating with respect to x,

Again differentiating with respect to x,

For maxima or minima,

P'(x) = 0,

For x = 185,

P''(x) = negative,

Hence, for maximising the profit 185 units must be produced.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

step-by-step explanation:

consider an orange peel: if you want to try and lay it flat, you have to stretch it, we know exactly how things are being stretched or squashed at any given point. we have many different map projections because each has different patterns of some projections can even preserve certain features of the earth without