Answer :
Answer:
[tex](11\frac{1}{3},20)[/tex]
Step-by-step explanation:
If the Number of Sales of x units, N(x) is defined by the function:
[tex]N(x)=-5x^3+235x^2-3400x+20000,10\leq x\leq 40[/tex]
[tex]N'(x)=-15x^2+470x-3400\\When -15x^2+470x-3400=0\\x=20, x=11\frac{1}{3}[/tex]
Next, we text the critical points and the end points of the interval to see where the derivative is increasing.
[tex]N'(10)=-200\\N'(11\frac{1}{3})=0\\N'(19)=115\\N'(20)=0\\N'(40)=-8600[/tex]
Thus, the rate of change of sales [tex]N'(x)[/tex] is increasing in the interval [tex](11\frac{1}{3},20)[/tex] on 10≤x≤40.
The rate of change is how much the company sales changes after spending
The rate of change increases at: (10, 20)
The function is given as:
[tex]\mathbf{f(x) = -5x^3 +235x^2 - 3400x + 20000}[/tex]
Differentiate, to calculate the rate of change
[tex]\mathbf{f'(x) = -15x^2 +470x- 3400}[/tex]
Set to 0
[tex]\mathbf{ -15x^2 +470x- 3400 = 0}[/tex]
Divide through by 5
[tex]\mathbf{ -3x^2 +94x- 680 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = \{20,11.3\}}[/tex]
The domain is given as: [tex]\mathbf{10 \le x \le 40}[/tex]
When x = 10, we have:
[tex]\mathbf{f'(10) = -15(10)^2 +470(10)- 3400 = -200}[/tex]
When x = 11.3, we have:
[tex]\mathbf{f'(11.3) = -15(11.3)^2 +470(11.3)- 3400 = -4.35}[/tex]
When x = 20, we have:
[tex]\mathbf{f'(20) = -15(20)^2 +470(20)- 3400 = 0}[/tex]
When x = 40, we have:
[tex]\mathbf{f'(40) = -15(40)^2 +470(40)- 3400 = -8600}[/tex]
The value of f'(x) increases at x = 10 to 20.
Hence, the rate of change increases at: (10, 20)
Read more about rates at:
https://brainly.com/question/18576593