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the population density d (in people/mi²) in a large city is related to …

Question

the population density d (in people/mi²) in a large city is related to the distance x (in miles) from the center of the city by (d=\frac{4600x}{x^{2}+15}). (a) what happens to the density as the distance from the center of the city changes from 20 miles to 25 miles? the density decreases. the density does not change. the density increases. (b) what eventually happens to the density? the density does not change. the density gets closer to ∞. the density gets closer to 10. the density gets closer to 0. (c) in what areas of the city does the population density exceed 400 people/mi²? areas between 1.5 and 10 miles from the center of the city areas more than 10 miles away from the center of the city areas more than 1.5 miles away from the center of the city areas within 10 miles of the center of the city

Explanation:

Step1: Calculate density at $x = 20$

Substitute $x = 20$ into $D=\frac{4600x}{x^{2}+15}$:
$D_1=\frac{4600\times20}{20^{2}+15}=\frac{92000}{400 + 15}=\frac{92000}{415}\approx221.69$

Step2: Calculate density at $x = 25$

Substitute $x = 25$ into $D=\frac{4600x}{x^{2}+15}$:
$D_2=\frac{4600\times25}{25^{2}+15}=\frac{115000}{625+15}=\frac{115000}{640}\approx179.69$
Since $D_1>D_2$, the density decreases as the distance changes from 20 miles to 25 miles.

Step3: Analyze the limit as $x\to\infty$

We find $\lim_{x\to\infty}\frac{4600x}{x^{2}+15}$. Divide both numerator and denominator by $x^{2}$:
$\lim_{x\to\infty}\frac{\frac{4600x}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{15}{x^{2}}}=\lim_{x\to\infty}\frac{\frac{4600}{x}}{1 + \frac{15}{x^{2}}}=0$

Step4: Solve the inequality $D>400$

Set $\frac{4600x}{x^{2}+15}>400$.
Multiply both sides by $x^{2}+15$ (since $x^{2}+15>0$ for all real $x$) to get $4600x>400(x^{2}+15)$.
Expand to $4600x>400x^{2}+6000$.
Rearrange to $400x^{2}-4600x + 6000<0$.
Divide by 200: $2x^{2}-23x + 30<0$.
Factor: $(2x - 3)(x - 10)<0$.
The roots of the equation $(2x - 3)(x - 10)=0$ are $x = 1.5$ and $x = 10$.
The solution of the inequality is $1.5

Answer:

(a) The density decreases.
(b) The density gets closer to 0.
(c) areas between 1.5 and 10 miles from the center of the city