Question

The function

Answer 1

Corrected Question

The function is: $R(x)=80+7.440\sqrt{x} , 0\leq x\leq 15000$

Answer:

(a)$48.02

(b)$656,299.92

Step-by-step explanation:

(a) We are to determine the change in monthly revenue if the monthly expenditure, x is raised from its current level of 6000 to 6001.

$R(x)=80+7.440\sqrt{x}\\R'(x)=\dfrac{d}{dx} (80+7.440\cdot x^{1/2})\\=7.440 \times \frac{1}{2} \times x^{\frac{1}{2}-1}\\=3.72\times x^{-\frac{1}{2}}\\\\R'(x)=\dfrac{3.72}{\sqrt{x} }$

Therefore, the expected change in monthly revenue

$R'(6000)=\dfrac{3.72}{\sqrt{6000} }= 0.04802 (thousands)$

=$48.02

(b)When the amount spent on advertising is $6000

$Revenue, R(6000)=80+7.440\sqrt{6000}\\=\ 656.29992 (in thousands)\\=\ 656,299.92$