THIS IS THE COMPLETE QUESTION BELOW
The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.
Answer
$168.27
Step by step Explanation
Given p=90000/400+3x
With the limits of 40 to 50
Then we need the integral in the form below to find the average price
1/(g-d)∫ⁿₐf(x)dx
Where n= 40 and a= 50, then if we substitute p and the limits then we integrate
1/(50-40)∫⁵⁰₄₀(90000/400+3x)
1/10∫⁵⁰₄₀(90000/400+3x)
If we perform some factorization we have
90000/(10)(3)∫3dx/(400+3x)
3000[ln400+3x]₄₀⁵⁰
Then let substitute the upper and lower limits we have
3000[ln400+3(50)]-ln[400+3(40]
30000[ln550-ln520]
3000[6.3099×6.254]
3000[0.056]
=168.27
the average price p on the interval 40 ≤ x ≤ 50 is
=$168.27
Answer: 64 years
Step-by-step explanation:
Let assume the dealer sold the bottle now for $P, then invested that money at 5% interest. The return would be:
R1 = P(1.05)^t,
This means that after t years, the dealer would have the total amount of:
$P×1.05^t.
If the dealer prefer to wait for t years from now to sell the bottle of wine, then he will get the return of:
R2 = $P(1 + 20).
The value of t which will make both returns equal, will be;
R1 = R2.
P×1.05^t = P(1+20)
P will cancel out
1.05^t = 21
Log both sides
Log1.05^t = Log21
tLog1.05 = Log21
t = Log21/Log1.05
t = 64 years
The best time to sell the wine is therefore 64years from now.
Answer:
Im not good at these so maybe 105°? Sorry if its wrong
Step-by-step explanation:
