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olchik [2.2K]
2 years ago
11

What is the answer to -2t-6z-1=?

Mathematics
1 answer:
erik [133]2 years ago
4 0

Answer:

2t-6z-1

The what I got

I home it will help

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Elasticity of Demand The demand function for a certain brand of backpacks is
Trava [24]

Answer:

See explanation

Step-by-step explanation:

Solution:-

- The demand function of a certain brand is given as price P a function of x quantity of goods ( in hundred ) demanded per month. The relation is:

                           P ( x ) = 50 Ln ( 50 / x ).

- The point price elasticity ( E ) of demand is given by:

                           E = \frac{P}{x}*\frac{dP}{dx}  

- Where, dP / dx : is the rate of change of price ( P ) with each hundred unit of good ( x ) is demanded.

- To determine the " dP / dx " by taking the first derivative of the given relation:

                          P ( x ) = 50 Ln ( 50 / x ).

                          d P ( x ) / dx = [ 50*x / 50 ] * [ -1*50 / x^2 ]

                                              = - 50 / x

- Hence the point price elasticity of demand is given by:

                          E = - ( P / x ) * ( 50 / x )

                          E = -50*P / x^2    

- For an inelastic demand, ! E ! is < 1:

                          ! -50*P / x^2 ! < 1

                          50*P / x^2 < 1

                          P < x^2 / 50

- For an unitary demand, ! E ! is =  1:

                          ! -50*P / x^2 !  = 1

                          50*P / x^2 = 1

                          P = x^2 / 50

- For an inelastic demand, ! E ! is > 1:

                          ! -50*P / x^2 ! > 1

                          50*P / x^2 > 1

                          P > x^2 / 50

2)

If the unit price is increased slightly from $50, will the revenue increase or decrease?

- We see from the calculated demand sensitivity d P / dt:

                          d P ( x ) / dx = - 50 / x

- We see that as P increases the from P = $50, the quantity of goods demanded would be:

                          50 = 50 ln(50/x)

                           1 = Ln ( 50 / x )

                           50/x = e

                           x = 50 / e

Then,

                          d P ( x ) / dx = - 50 / ( 50 / e )

                          d P ( x ) / dx = - e

- We see that if price slightly increases from $ 50 then the quantity demanded would decrease by e (hundreds ) goods.

- The decrease in the quantity demanded is higher than the increase in price. The revenue is given by the product of price P ( x ) and x:

                Revenue R ( x ) = P ( x ) * x

                                = 50*x*ln(50/x)

Then the product of price and quantity goods also decreases; hence, revenue decreases.

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What is the correct maximum or minimum for the function f(x) = -|x-2| + 1
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