Applications of maxima?

Question

1. For a rectangle with area 289 to have the smallest perimeter, what dimensions should it have?

2. The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. The following is a possible model, where q is the reduction in emissions (in pounds of pollutant per day) and C is the daily cost to the firm (in dollars) of this reduction.

c(q)=4900+90q^2
*What level of reduction corresponds to the lowest average cost per pound of pollutant? (Give the answer to two decimal places.)
*What would be the resulting average cost to the nearest dollar?

3.Hercules Films is deciding on the price of the video release of its new film. Its marketing people estimate that at a price of p dollars, it can sell a total of q copies. What price will bring in the greatest revenue?
q=220,000-10000p copies

Answer 1

1. We all know it will be a square, so 17×17. But if you want to prove it, or more accurately if your teacher wants you to prove it (they're like that, after all), it goes like this: Let the width be x and the length 289/x. Then the perimeter is 2x + 578/x, which has derivative 2 - 578/x^2, which will give a maximum when x^2 = 289, i.e. x = 17 - big surprise, eh?

2. c(q) = 4900 + 90q^2
cost per pound = c(q) / q = 4900/q + 90q = f(q)
f'(q) = 90 - 4900/q^2 = 0 when q^2 = 4900/90 => q = 7√10 / 3 ≈ 7.379 lb/day.
Average daily cost per pound is 4900/7.379 + 90(7.379) = $1328 (nearest dollar).

3. Revenue = r = pq = 220,000p - 10,000p^2
This is a quadratic with negative p^2 coefficient, so we know it has a maximum at p = -220,000 / (-20,000) = $11.

Answer 2

1. For a rectangle with area 289 to have the smallest perimeter, what dimensions should it have?

Solution
17 by 17

Answer 3

I'll show you number 1:

As always, l is length, w is width, p is perimeter,
and A is area.

So we have A=lw=289 and p=2l+2w

So l=289/w and we have

p=(578/w)+2w
=(2w^2+578)/w

Now we minimize p

p'=(w(4w)-2w^2-578)/w^2

=(2w^2-578)/w^2

Set this equal to 0 and we get

w^2=289

w=17

Using the first derivative test, we can see that p' changes from - to + at w=17, so this is where p is minimized.

Since l=289/w

We have the minimal dimensions as

l=w=17