Econ 407

Assignment: Inst. N. hash: Applications of derivatives: how to find the minimum and

maximum of  cost and revenue functions.

1.The cost of producing q items is given by C = 15q^(2/3) + 100q + 10,000.

Find where this cost function is increasing and where

it is decreasing. Sketch the graph of C

> C:=15*q^(2/3) + 100*q + 10000;

C := 15*q^(2/3)+100*q+10000

> plot(C,q = 0..40000);

As the plot of the cost curve shows, it is an increasing function of output--q. We can also take the first derivative and evaluate the first derivative for various values of q. But the rate of increase in the cost function decreases. In other words, it increases at a decreasing rate.

> dc:=diff(C,q);

dc := 10/q^(1/3)+100

> eval(dc, q=100.0);

102.1544347

> eval (dc, q= 1000.0);

101.0000000

> plot(dc, q=0...43000);

>

If a company sells q items per day, then the price of each item is given by

P = 10 - q/10

Find the range where the revenue function R(q) =qp is increasing and where it is

decreasing.

> TR:=(10.0-q/10)*q;

TR := (10.0-1/10*q)*q

> plot(TR, q=0..120);

As can be seen from the graph, TR increases until output reaches about 50 units, then it begins to drop off. Beyond q=100, TR becomes negative. We can find the maximum point on TR by taking its derivative and setting equal to zero and solving for q.

> MR:=diff(TR,q);

MR := -1/5*q+10.0

> solve(MR=0,q);

50.

>

3. The cost of producing q items is given by C(q) = 100 +50q + 0.2q^3. Find

dC/dq and d^2C/dq^2 at q = 0, 5, 50, and 75. Find the first and the second

derivative of the cost function. The second derivative is a useful concept and

will be discussed in class at a later date.

> C:= 100 +50*q + 0.2*q^3;

C := 100+50*q+.2*q^3

> dc:=diff(C,q);

dc := 50+.6*q^2

> d2c:=diff(dc,q);

d2c := 1.2*q

> plot(C, q=150..300);

>

4. The postage for shipping an item weighing q ounces is modeled by

P = Q(Q^2 +1)^(0.5)

Find the dP/dq and d^2P/dq2 when q = 1, 10 and 20.

> P:=(q)*(q^2 +1)^(0.5);

P := q*(q^2+1)^.5

> plot(P,q=200..1000);

> dp:= diff(P,q);

dp := (q^2+1)^.5+1.0*q^2/(q^2+1)^.5

> ddp:=diff(dp,q);

ddp := 3.0/(q^2+1)^.5*q-1.00*q^3/(q^2+1)^1.5

> eval(dp,q=1);

2.121320343

> eval(ddp,q=1);

1.767766953

> eval(dp=10);

(q^2+1)^.5+1.0*q^2/(q^2+1)^.5 = 10

> eval(ddp,q=10);

1.999926234

>

5. Suppose the daily profit (in dollars) from manufacturing q items is modeled

by P(q)= -0.01q^2 + 5q-400, where 0<q<300. Find the number of items q that

results in maximum profits.

Take the derivative of the profit function and set equal to zero and solve

for q.

6. Suppose the cost (in dollars) of manufacturing q items is modeled by

C =400+ 10q +0.01q^2. Find the value of q that minimizes the average cost

function. Solution: first compute AC function. This is done by dividing the

total cost function by output--q, as shown below.

> C:=400.0+ 10*q +0.01*q^2;

C := 400.0+10*q+.1e-1*q^2

> AC:=C/q;

AC := (400.0+10*q+.1e-1*q^2)/q

> plot(AC,q=100..350);

Looking at the graph, the function shows a minimum around q=200. To be more exact, we take the first derivative of the AC function with respect to q, and set equal to zero.

> dAC:=diff(AC,q);

dAC := (10+.2e-1*q)/q-(400.0+10*q+.1e-1*q^2)/q^2

> solve(dAC,q);

200.   -200.

So the solution is at q=200.