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Composite Functions: Application

The price per unit, p, for the product is p = 2000 – 10t, where t is the number of months past January 1995.

Example: The monthly demand, D, for a product, is

,000,000,5p

D where p is the price per unit of the product.

Write the monthly demand, D, as a function of t.

Compute (D p)(t) = D(p(t)).

Note, D is a function of p, D(p)D

p

(D p)(t) = .102000

000,000,5t

and p is a function of t.

tp(t)

Table of Contents

Composite Functions: Application

Slide 2

(D p)(t) =t102000

000,000,5

When will the monthly demand reach 6,250 units?

This is now a function of demand with respect to t, so can

.102000

000,000,5t

be relabeled, D(t) =

6250 = ,102000

000,000,5t

6250(2000 – 10t) = 5000000,

12500000 – 62500t = 5000000, - 62500t = - 7500000,

t = 120 months The monthly demand will reach 6,250 units in January 2005.

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Composite Functions: Application

Slide 3

Try: An observer on the ground is 300 feet away from the launching point of a balloon. The balloon is rising is rising at a rate of 10 feet per second. Let d = the distance (in feet) between the balloon

and the observer.Let t = the time elapsed (in seconds) since the

balloon was launched.Let x = the balloon's altitude (in feet).

300 feet

xdJot down the figures above and click to see the questions!

Table of Contents

Composite Functions: Application

(a) Express d as a function of x. Hint: Use the Pythagorean Theorem.

Slide 4

300 feet

xd

(b) Express x as a function of t.

(c) Express d as a function of t.

(d) Use the result found in (c) todetermine how long it takes from launching for the balloonto be 500 feet from theobserver.

22300 xxd

x(t) = 10t

22 )10(300 ttdtxd

It takes 40 seconds.

Table of Contents

Composite Functions: Application