Q1 Step 1

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise The supply function for a product is given by p = 30 + 100v 2x + 21, where x is the number of units supplied and p is the price in dollars. If the price is increasing at a rate of $1 per month, at what rate is the supply changing when the price per unit is $930? Step 1 The relationship between price p in dollars and the number of units x is given by p = 30 + 100 \\ 2x + 21. Rewrite the supply function in exponential form. p = 30 + 100 \\ 2x + 21 P = 30 + 100(2x + 21) 1/2 1/2 If the price is increasing at a rate of $1 per month, this is a rate of change of price p with respect to time t and dp/dto dp/dt = 1. The unknown is the rate at which the supply is changing when the price per unit is $930. This is a rate of change of x with respect to time t or dx/dt dx/dtStep 2 To find x , begin by finding the derivative of the supply function. Use the Power Rule and Chain Rule to find the derivative of p with respect to x for p = 30 + 100(2x + 21)1/2. Simplify, and then write in radical form. dp = 0 + 100 . (2x + 21)-1/2 . 30 X dx 2 dp 100 100 (2x + 21)-1/2 dx 100 dp 2x + 21 2x + 21 Step 3 dp dp dx The Chain Rule gives us the derivative of p with respect to t as - dt dx dt 100 Substitute - dp - = 1 and ap dx dt dx V 2x + 21 , and solve for . dt dp dx dt dx dt 100 100 dx 1 = dt V 2x + 21 44 X V2c + 21 dx dt 100Notice that - dx V 2x + 21 dt is a function in terms of x, so to evaluate the rate, we must find the value of x when the price is $930. 100 Using the demand function p = 30 + 100 v 2x + 21, replace p with 930 and solve for x. 930 = 30 + 100 \\ 2x + 21 900 V = 100\\ 2x + 21 X = V2x + 21 X = 2x + 21 X = 2x X =X