the Calculus is killing my grades

calculus problem-please someone help me out not the best at this p=D(x)=125e^-0.001x ; p=S(x)=25e^0.001x

The value of x at equilibrium is___________

Find the consumer's surplus and the producers' surplus at the equilibrium price level for the given price demand and price supply equations. Round all values to nearest integer.(Crying)

Re: the Calculus is killing my grades

In order to start this problem we must note that at equilibrium Demand=Supply and therefore D(X)=S(X)

$\displaystyle 125 e^{-0.001x} = 25 e^{0.001x}$

$\displaystyle e^{0.002x} = \frac{125}{25}$

$\displaystyle e^{0.002x}=5$

$\displaystyle 0.002x = ln(5)$

$\displaystyle x = \frac{ln(5)}{0.002}$

$\displaystyle x = 1.6094$

$\displaystyle D(1.6094) = 125e^{(-0.001(1.6094))} =124.80$

$\displaystyle S(1.6094) = 25 e^{(0.001(1.6094))} = 25.04$

Equilibrium demand and price $\displaystyle = q(e),p(e) = [ 124.80, 25.04]$

$\displaystyle p(e)q(e) = (124.80)(25.04) = 3,124.99$

CS$\displaystyle = 125 \int^1_0 e^{-0.001x} dx -p(e)q(e)$

$\displaystyle \int e^{-0.001x} dx = -e^{-0.001x} /.001 = -1,000e^{-0.001x}$

$\displaystyle F(x) = -125,000 e^{-0.001x}$

$\displaystyle F(124.80) = -125,000e^{(-0.001(124.80))} = -110334.177$

$\displaystyle F(0) = -125000$

$\displaystyle F(124.80)-F(0) = -110334.177 - (-125,000) = 14,665.82$

$\displaystyle 14,665.82 - p(e)q(e) = 14,665.82- 3,124.99 = 11540.83$

$\displaystyle CS = 11,540.83$

moving on to PS

$\displaystyle =p(e)q(e) - 25 \int e^{0.001x} dx$

$\displaystyle \int e^{0.001x} dx = 1000e^{0.001x}$

$\displaystyle 25,000 e^{0.001x}$

$\displaystyle Let F(x) = 25,000 e^{0.001x}$

$\displaystyle F(25.04) = 25633.90$

$\displaystyle F(0) = 25000$

$\displaystyle F(25)-F(0) = 25633.90 - 25000 = 633.90$

$\displaystyle p(e)q(e) = 3,124.99 -633.90 = 2,491.09$

$\displaystyle PS = 2,491.09$

you have posted similar problems now try those and see if you can follow the general procedure i have demonstrated

Re: the Calculus is killing my grades

thanks for the help. Got 2 more for you math brains.

A contaminated lake is treated with bactericide.The rate of increase in harmful bacteria t days after the treatment is given by the following function where N(t) is the number of bacteria per milliliter of water. Since dN/dt is negative, the count of harmful bacteria is decreasing.

dN/dt=-2000t/(1+t^2) 0=too or less then t =too less then 8

(A) Find the minimum value of dN/dt

The minimum value of dN/dt is________bacteria/ml per day

#2-Find the indefinite integal

/ x^5e^x^6 dx=

use integers or factions for any numbers in the expression. use C as the arbitrary constant