Question

Previous Answers 12 points CraudColAlg5 2AFL3.007 My Notes Ask Your Teacher The following illustrates an application of optimization using parabolas. A study found that the cost C per pupil of operating a high school in a certain country depends on the number n of students enrolled. The cost is given by C=65,000-500n+ n? oolars per pupi What enrollment produces the minimum cost per pupil? n250 students What is the minimum cost? 25000 Need Help? Read Talk to a Tutor I Previous Arswers CraudColAlgs 24 58.002 10 O1/1 points My Notes Ask Your Teacher Use the crossing-graphs method to solve the given equation. (Round your answer to two decimal places.) 2-4/59 x=194 Need Help? Read Watch Talk to a Tutor Previous Arewers 11. 01 points CraudColAigs 2.1.58.014 My NotesO Ask Your Teacher It is a fact that the function ( +x)x has a limiting value. Use a table of values to estimate the limiting value. (Round your answer to two decimal places.) 272 Read Need Help? Talk to a Tuter Previous Arswers CraudColAlg5 21.58.021 12. 02 points My Notes Ask Your Teacher Suppose the function f(x) x- 2x + 43 describes a physical situation that makes sense only for whole numbers between 0 and 20 (such as family expense as a function of the number of children). For what value of x does f reach a minimum, and what is the minimum value? (Suggestion: We suggest a table starting at 0 with a table increment of 1.) xvalue minimum value of Need Help? Read Talk to a Tuter

Answer 1

9) Given function is : $C=65000-500n+n^2$

Putting n = 250 we get,

$C=65000-500*250+250^2$

i.e., $C=65000-125000+62500$

i.e., $C=127500-125000$

i.e., $C=2500$

Therefore, the minimum cost is $2500.

11) Let $y=(1+x)^{1/x}$

Taking log_e on both sides we get,

$\ln y=\ln (1+x)^{1/x}$

i.e., $\ln y=\frac{1}{x}\ln (1+x)$

i.e., $\ln y=\frac{\ln (1+x)}{x}$

Then, $\lim_{x\to \infty}\ln y=\lim_{x\to \infty}\frac{\ln (1+x)}{x}$

i.e., $\lim_{x\to \infty}\ln y=\lim_{x\to \infty}\frac{1/(1+x)}{1}$    [Using L'hospital rule]

i.e., $\lim_{x\to \infty}\ln y=\lim_{x\to \infty}\frac{1}{1+x}$

i.e., $\lim_{x\to \infty}\ln y=\lim_{x\to \infty}\frac{0}{1}$ [Using L'hospital rule]

i.e., $\lim_{x\to \infty}\ln y=0$

i.e., $\ln\left (\lim_{x\to \infty}y \right )=0$

i.e., $\left (\lim_{x\to \infty}y \right )=e^0$

i.e., $\lim_{x\to \infty}y=1$

i.e., $\lim_{x\to \infty}(1+x)^{1/x}=1$

Therefore, limiting value of the given function is 1.

12) Given function is : $f(x)=x^2-2x+43$

For minimum value of f(x) we must have $\frac{d}{dx}f(x)=0$ .

i.e., $\frac{d}{dx}(x^2-2x+43)=0$

i.e., $2x-2=0$

i.e., $2x=2$

i.e., $x=1$

Therefore, required value of x is 1.

Putting x = 1 in the given function we get,

$f(1)=1^2-2*1+43$

i.e., $f(1)=1-2+43$

i.e., $f(1)=42$

Therefore, minimum value of f(x) is 42.