MATH120 Lecture Notes 31 - Applications of maxima and minima PDF

Preview

Summary

Download MATH120 Lecture Notes 31 - Applications of maxima and minima PDF

Description

112

31 APPLICATIONS OF MAXIMA AND MINIMA

31

Applications of Maxima and Minima

Many situations arise in practice when we want to maximize or minimize a certain quantity. The general rule is that we express this quantity as a function of a certain other quantity which is regarded as a variable, for example, the cost function is a function with variable x, which stands for the number of items to be produced. The extremum of the function in general occurs at a critical point of the function, and we can use the second derivative test to confirm whether the critical point is a local minimum or local maximum point. We will explain the use of this general method through various examples. Example 1 (Maximizing Profits) A small manufacturing firm can sell all the items it can produce at a price of $6 each. The cost of producing x items per week (in dollars) is C(x) = 1000 + 6x − 0.003x2 + 10−6 x3 . What value of x should be selected in order to maximize the profit? Solution The revenue from selling x items at $6 each is R(x) = 6x dollars. Therefore the profit per week is P (x) = R(x) − C(x)

= 6x − (1000 + 6x − 0.003x2 + 10−6 x4 ) = −1000 + 0.003x2 − 10−6 x3

To find x which maximizes P (x), we first find the critical points of P (x). P ′ (x) = (−1000 + 0.003x2 − 10−6 x3 )′ = 0.006x − 10−6 3x2

Letting P ′ (x) = 0 we obtain 0.006x − 10−6 · 3x2 = 0

x(0.006 − 10−6 · 3x) = 0 It gives x = 0 or x =

0.006 10−6 ·3

= 2000.

Thus we have two critical points x = 0 and x = 2000. We now use the second derivative test to determine which critical point is a maximum point. P ′′ (x) = (0.006x − 10−6 · 3x2 )′

113 = 0.006 − 10−6 · 6x P ′′ (10) = 0.006 > 0, P ′′ (2000) = 0.006 − 10−6 · 6(2000)

= 0.006 − 0.012 = −0.006 < 0

Therefore x = 2000 maximizes P (x), that is x = 2000 should be selected in order to maximize the profits. ✷ Example 2 (Price Decision) The cost of producing x items per week is C(x) = 1000 + 6x − 0.003x2 + 10−6 x3 For the particular item in question, the price at which x items can be sold per week is given by the demand equation p = 12 − 0.0015x Determine the price and volume of sales at which the profit is maximum. Solution The revenue per week is R(x) = px = (12 − 0.0015x)x = 12x − 0.0015x2 The profit is therefore given by P (x) = R(x) − C(x)

= (12x − 0.0015x2 ) − (1000 + 6x − 0.003x2 + 10−6 x3 )

= −1000 + 6x + 0.0015x2 − 10−6 x3

To find when P (x) has a maximum, we first set P ′ (x) = 0 to find the critical points. P ′ (x) = 6 + 0.003x − (3 × 10−6 )x2 = 0 To find x, we use the quadratic formula and obtain p −0.003 ± 0.0032 − 4 × 6 × (−3 × 10−6 ) x = 2(−3 × 10−6 ) √ −0.003 ± 0.00009 + 72 × 10−6 = −6 × 10−6 = 2000 or − 1000 The negative solution has no practical significance, and when x = 2000,

P ′′(x) = 0.003 − 6 × 10−6 x

= 0.003 − 6 × 10−6 × 2000

= 0.003 − 0.012

= −0.009 < 0

114

31 APPLICATIONS OF MAXIMA AND MINIMA Therefore P (x) has a maximum when x = 2000. By the demand equation, p = 12 − 0.0015x

= 12 − 0.0015 × 2000

= 9

Thus the sales volume of 2000 items per week gives the maximum profit, and the price per item corresponding to this value of x is 9 (dollars). ✷ Example 3 (Minimizing Cost) A tank is to be constructed with a horizontal, square base and vertical, rectangular sides. There is no top. The tank must hold 4 cubic meters of water. The material of which the tank is to be constructed costs $10 per square meter. What dimensions for the tank minimizes the cost of material? Solution Let x be the length of the sides and y the height of the tank. Then the square base has area x2 , and the four rectangular sides each has area xy.

The total area of the material used is A = x2 + 4xy The volume of the tank is V = x2 y but by requirement, V = 4. Thus x2 y = 4 which gives y = x42 . Substituting this into the expression of A we obtain µ

4 A = x + 4xy = x + 4x x2 2

2

¶

= x2 +

16 x

Now we can find for what value of x the area A has a minimum, and therefore the cost is minimized. dA = dx

¶′ µ 16 16 2 = 2x − 2 x + x x

115 To find critical points of A we let 2x −

16 x2

= 0,

2x =

16 , x2

dA dx

= 0, i.e.

2x3 = 16, x3 = 8,

x=2

There is one critical point x = 2. To use the second derivative test, we calculate µ ¶ 16 ′′ d2 A 48 = 2x − 2 = 2+ 3 dx2 x x and at x = 2, 48 48 d2 A >0 = 2+ =2+ 3 2 8 (2) dx Therefore A has a minimum when x = 2. From y = when x = 2.

4 x2

we deduce y =

4 22

=1

Thus the tank should have the sides each 2 meters in length and 1 meter in height to minimize the cost. ✷...