# COMPOSITION OF FUNCTION WORD PROBLEMS

## Composition of Functions

Here we can find problems that how we are finding composition of two functions from word problems.

To find composition of two functions f and g, we have to follow the procedure given below.

Step 1 :

In (f∘g) (x),

Write f and remove the composition sign. Inside the bracket put the function g(x). So, we will get

(f∘g) (x) = f[g(x)]

Step 2 :

In the place of g(x), put the respective function.

Step 3 :

Now the function g(x) is like a input for the function f(x). So, apply the function g(x) in the place of x in the function f(x).

Problem 1 :

The surface area S (in square meters) of a hot - air balloon is given by

S(r) = 4πr2

Where r is the radius of the balloon (in meters). If the radius r is increasing with time t (in seconds) according to the formula

r(t) = 2/3t3, t ≥ 0

find the surface area S of the balloon as a function of the time t.

Solution :

S = surface area

Surface area is S(r) = 4πr2 --- (1)

Where r is the radius of the balloon.

If the radius r is increasing with time t (in seconds).

r(t) = 2/3t3

r(t) = 2/3t3 substitute in equation (1).

S(r(t)) = 4π(r(t))2

Problem 2 :

The volume V (in cubic meters) of the hot – air balloon described in Problem 65 is given by

V(r) = 4/3 πr3

If the radius r is the same function of t as in Problem 65, find the volume V as a function of the time t.

Solution :

Problem 3 :

The number N of cars produced at a certain factory in one day after t hours of operation is given by

N(t) = 100t – 5t2, 0 ≤ t ≤ 10

If the cost C (in dollars) of producing N cars is

C(N) = 15,000 + 8000N

find the cost C as a function of the time t of operation of the factory.

Solution :

N(t) = 100t – 5t2

C(N) = 15,000 + 8000N

C(N(t)) = 15,000 + 8000N(t)

= 15,000 + 8000(100t - 5t2)

C(N(t)) = 15,000 + 800, 000t - 40,000t2

Problem 4 :

The spread of oil leaking from a tanker is in the shape of a circle. If the radius r (in feet) of the spread after t hours is

r(t) = 200 √t

find the area A of the oil slick as a function of the time t.

Solution :

area A = πr2

Given, r(t) = 200 √t

A(t) = π(200√t)2

= π(200)2(t1/2)2

A(t) = 40000πt

Problem 5 :

The price p, in dollars, of a certain product and the quantity x sold obey the demand equation

p = -1/4 x + 100   0 ≤ x ≤ 400

Suppose that the cost C, in dollars, of producing x units is

C = √x/25 + 600

Assuming that all items produced are sold, find the cost C as a function of the price p.

[Hint : Solve for x in the demand equation and then form the composite]

Solution :

First we need to solve for x in the demand equation :

p = -1/4 x + 100

Add 1/4 x on both sodes.

p + 1/4 x = 100

Subtract p on each sides.

1/4 x = 100 - p

Multiply 4 on each sides.

x = 4(100 - p)

x = 400 - 4p

Problem 6 :

Cost of a Commodity : The price p, in dollars, of a certain commodity and the quantity x sold obey the demand equation

p = -1/5 x + 200   0 ≤ x ≤ 1000

Suppose that the cost C, in dollars, of producing x units is

C = √x/10 + 400

Assuming that all items produced are sold, find the cost C as a function of the price p.

Solution :

Given, p = -1/5 x + 200

Add 1/5 x on both sodes.

p + 1/5 x = 200

Subtract p on each sides.

1/5 x = 200 - p

Multiply 5 on each sides.

x = 5(200 - p)

x = 1000 - 5p

Problem 7 :

The volume V of a right circular cylinder of height h and radius r is V = πr2h. If the height is twice the radius, express the volume V as a function of r.

Solution :

V = πr2h

The height is twice the radius.

So, V(r) = 2r(πr2)

V(r) = 2πr3

Problem 8 :

The volume V of a right circular cone is

V = 1/3 πr2h

If If the height is twice the radius, express the volume V as a function of r.

Solution :

V = 1/3 πr2h

The height is twice the radius.

So, V(r) = 2r(1/3 πr2)

V(r) = 2/3 πr

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