S(q) = 10 15e^(0.03q) ; q0 = 3 units

Question

p= S(q) is the price (dollars per unit) at which q units of a particular commodity will be supplied to the market by producers, and q0 is a specified level of production. Find the price p0 = S(q0) at which q0 units will be supplied and compute the corresponding producer’s surplus PS.

S(q) = 10 15e^(0.03q) ; q0 = 3 units

Answer 1

Question: Note: "+" added

S(q) = 10+15e^(0.03q); q0 = 3 units

p= S(q) is the price (dollars per unit) at which q units of a particular commodity will be supplied to the market by producers, and q0 is a specified level of production.

Find the price p0 = S(q0) at which q0 units will be supplied and compute the corresponding producer’s surplus PS.

Answer:

Hi, I attempt the questions that are left unanswered for a long time. This usually means I must research to find the answer, because if it was easy It would have been answered already.

At first, I though straight substitution and then evaluate:

S(q) = 10+15e^(0.03q); q0 = 3 units

S(q) = 10+15e^(0.03*3)

S(q) = 10+15e^(0.09)

How do you take e to a power of 0.09?  AND  What is e?

How do you take e to a power of 0.09?

S(q) = 10+15e^(9/100) ==> S(q) = 10 +15*[100sqrt (e^9)]

What is e?

Mathmatically, e occurs in population growth problems, in the statistics problems, in the sequences and series problems.

It crops up when you start Calculus.

e = 2.718281828459045..., the Base of Natural Logarithms

OR

lim as n->infinity (1+1/n)^n = e

"the limit as n goes to infinity of the quantity one plus one over n to the nth power is e"  [lim as n->infinity (1+1/n)^n = e]

Interesting fact: e^(i*Pi) + 1 = 0

Unique item: if you take the derivative of e^x, you get e^x.

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I have to go, but if I think of anything else, I'll update this.