Respuesta :
Answer:
a) x ≥ 0
b) 0
c) x = 0 ( min ) , x = 5.192 ( max )
d) Q = 0 ( min ) , Q = 0.67357 ( 1000 ) units
e) 0 < x < 5.192 ( increasing ) , x > 5.192 ( decreasing )
Step-by-step explanation:
Solution:-
- The thousand units of product [ Q(x) ] sold is modeled by the following equation:
[tex]Q(x) = \frac{7x}{27+x^2}[/tex]
Where,
x: Is the amount of cedis ( thousands ) invested
- We are to determine the domain of the function. We need to realize that the amount of money spent can only be a positive value. Hence, x ≥ 0.
- Next we check for any "discontinuity" of the function [ Q(x) ]. The discontinuity may exist for rational fractions, natural logs, and radicals.
- The defined function is the form of " rational fraction ". For such functions we equate the denominator to zero and solve for any " real values " of the independent variable ( x ):
[tex]27 + x^2 = 0\\\\ x = i\sqrt{27} = 5.1962i \\[/tex]
- The roots are imaginary; hence, the discontinuity does not exist and the [ Q ( x ) ] is valid for all real positive values of x. the domain would be:
domain: [tex]x\geq 0[/tex]
- To determine the amount of product sold when no money is spent on the marketing of the product. We simply plug ( x = 0 ) in the given relation:
[tex]Q( 0 ) = \frac{0}{27} = 0[/tex]
Answer: No unit of product is sold without the marketing expense
- To determine the amount of money spent on marketing that would lead to maximum and minimum number of product to be sold. We are asked to find the critical values of the given function.
- To determine the critical values we have to determine the first derivative of the given function:
[tex]Q'(x) = \frac{7(27-x^2)}{(27 + x^2)^2}[/tex]
[tex]Q''(x) = \frac{-14x(-x^2 + 81 )}{(x^2 + 27)^3}[/tex]
- The critical values conditions are:
[tex]Q'(x) = 0[/tex]
[tex]\frac{7(27-x^2)}{(27 + x^2)^2} = 0\\\\7(27-x^2) = 0\\\\x = \sqrt{27} = +5.1962 \\\\[/tex]
- The other critical value is defined by the domain of the function. i.e x = 0. Where, Q ( x ) = 0.
- Input the critical values x = 5.1962 and x = 0 into the second derivative and determine the nature of the critical points as follows:
[tex]Q''(5.192) = \frac{-14*5.1962( -27 + 81 )}{( 27 + 27 )^3} < 0\\\\Q''(0) = \frac{0}{27} = 0 \geq 0[/tex]
- The critical value x = 5192 is the maximizing expenditure while x = 0 is the minimizing expenditure.
- The corresponding maximum and minimum number of unit sold can be determined by plugging the critical values in the given relation [ Q(x) ]:
[tex]Q ( 5.192 ) = \frac{7*\sqrt{27} }{27+27} \\\\Q ( 5.192 ) = 0.67357 ( thou) \\\\Q ( 0 ) = \frac{7*0 }{0+27} \\\\Q ( 0 ) =0 ( thou) \\[/tex]
Answer: The maximum units sold would be 673.57. And the minimum unit sold would 0.
- We will determine the nature of function [ increasing or decreasing ] by finding a range of value for which Q'(x) > 0 or Q'(x) < 0.
- We use the domain between the critical points. [ 0 , 5.192 ]
[tex]Q'(x) = \frac{7x(x^2 - 27 )}{(27+x^2)^2} > 0[/tex] ( increasing )
- The next domain is x > 5.192.
[tex]Q'(x) = \frac{7x(x^2 - 27 )}{(27+x^2)^2} < 0[/tex] ( decreasing )
Answer: the number of unit sold increases from 0 < x < 5.192 and decreases when expenditure is x > 5.192
