# Math Help - Review problems that confuse me

1. ## Review problems that confuse me

First of all, I have a calculus final coming up and I couldn't even do some problems that were apparently basic. I would really appreciate it if someone could break down each question and try to explain to me how they got to the answer. I really don't want to fail this final.

1) The farmer plans to fence a regular pasture adjacent to a river. The pasture must contain 2,000,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?

2) A population of 500 bacteria is introduced into a culture and grows in number according to the equation p(t)=500 (1+4t/50+t^2) where t is measured in hours. Find the rate at which the population is growing at t=2.

3) find the horizontal and vertical asymptote of y=2x/x-3

Find the limit.

lim 2x/x-3
x>3+

lim 2x/x-3
x>3-

lim 2x/x-3
x>+infinity

lim 2x/x-3
x>-infinity

lim 3x^3-2x^2+4
x>1

lim 2x^2-x-3/x+1
x>-1

lim 2x-3/x+5
x>-5

lim sin x/5x
x>0

Prove that the lim f(x) = 2 and lim f(x)=infinity are approaching infinity.
x>5 x>0

2. Here is some help with the first three problems. Mine in red.

1) The farmer plans to fence a regular pasture adjacent to a river. The pasture must contain 2,000,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?

Draw a diagram. Called the sides x, x, and y. The length is then 2x+y (excluding the river side). Restriction xy = 2,000,000. Solve for x from the restriction, then plug into the perimeter function:

$x = \frac{2000000}{y}$
$P(y) = \frac{4000000}{y} + y$
Use calculus to optimize this function. Find the first derivative, then critical numbers, then show that you have minimized the perimeter.

2) A population of 500 bacteria is introduced into a culture and grows in number according to the equation p(t)=500 (1+4t/50+t^2) where t is measured in hours. Find the rate at which the population is growing at t=2.

Find the first derivative using the quotient rule, then plug in t=2. The value you get is a rate measured by the units bacteria/hour.

3) find the horizontal and vertical asymptote of y=2x/x-3

Horizontal: The degrees of the numerator and denominator are equal, therefore the horizontal asymptote is the ratio of the leading coefficient $y= \frac{2}{1} = 2$
Vertical: x values that make ONLY the denominator 0, so it's $x=3$

Good luck on the final!

3. Thank you very much for you explanations! They were very helpful!

On #2, I got 31.58. Could anyone verify my answer?

4. Originally Posted by Rainy2Day
Thank you very much for you explanations! They were very helpful!

On #2, I got 31.58. Could anyone verify my answer?
No problem. You're welcome.

For number 2, I get 30.864, verified with a graphing calculator and Mathematica's Derivative Calculator: Step-by-Step Derivatives

Good luck!

5. Originally Posted by Rainy2Day
First of all, I have a calculus final coming up and I couldn't even do some problems that were apparently basic. I would really appreciate it if someone could break down each question and try to explain to me how they got to the answer. I really don't want to fail this final.

1) The farmer plans to fence a regular pasture adjacent to a river. The pasture must contain 2,000,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?
Let x is length of the rectangular field, parallel to river flow.
Let y be the width of the rectangular field.

Area = xy = 2 000 000

$y = \frac{2000000}{x}$ ..............(1)

Length of fencing = Perimeter (P) around three sides of rectangular garden

P = x + 2y

$P = x + 2\left(\frac{2000000}{x}\right)$

$P = x + \frac{4000000}{x}$

Now, differentiate P

$P' = 1 - \frac{4000000}{x^2}$

for Max or Min, P' = 0

$
1 - \frac{4000000}{x^2}=0
$

$x = 2000$

$y = \frac{2000000}{x}$

$y = \frac{2000000}{2000}=1000$

Length of fencing = 2000 m and width of fencing = 1000 m

$P'' = 0 + \frac{8000000}{x^3}$

$P'' = + \frac{8000000}{2000^3}= +$

so, it shows fencing is minimum.