Need a help w/word problem

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juliettusa

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I don't know how to make an equation to solve it. The same problem with the next problem=(( Please help

Two hundred and forty meters of fencing is available to enclose a rectangular playground. What should be the dimensions of the playground to maximize the area?


A video rental service has 1000 subscribers, each of whom pays $15 per month. On the basis of survey, they believe that for each decrease of #0.25 in the montly rate, they could obtain 20 additional subscribers.At what rate will max revenue be obtained and how many subscribers will there be at that rate?
 
Hmm, interesting. Is this an algebra or calc class? These are optimization problems.
I can solve this by calculus. (the answer is always a square) This can be solved just using algebra, but it takes quite an understanding of concepts

This is how you set it up. P = perimeter (given), A = area

for a rectangle, P = 2x + 2y and A = xy

x being dimensions of one side of the rectangle and y being the other

Technically, you need to set up an equation in one variable and then set the derivative to zero, which will give you the maximum area (does this sound familiar?)

see if you can get to

A/y + y = 120
 
juliettusa said:
I don't know how to make an equation to solve it. The same problem with the next problem=(( Please help

Two hundred and forty meters of fencing is available to enclose a rectangular playground. What should be the dimensions of the playground to maximize the area?


A video rental service has 1000 subscribers, each of whom pays $15 per month. On the basis of survey, they believe that for each decrease of #0.25 in the montly rate, they could obtain 20 additional subscribers.At what rate will max revenue be obtained and how many subscribers will there be at that rate?
Click to expand...

For your first problem, the perimeter of the playground must be 240 m, since that is all the fencing available. Let x = length of the playground, and let y = width of the playground. For a rectangle,

perimeter = 2*length + 2 * width

240 = 2x + 2y
You can divide both sides of this equation by 2 to get smaller numbers:

120 = x + y

Now...suppose we solve this equation for y in terms of x:

120 - x = y

We're concerned with the AREA of the playground...we'd like that to be the largest possible (maximum) area.

Area = length * width
Area = x*y
Substitute (120 - x) for y, and we have

Area = x(120 - x), or
Area = 120x - x[sup:2btiyuzu]2[/sup:2btiyuzu], or
Area = -x[sup:2btiyuzu]2[/sup:2btiyuzu] + 120x

Do you see that you have the equation of a parabola? This parabola opens DOWNWARD because the coefficient of x[sup:2btiyuzu]2[/sup:2btiyuzu] is negative. A downward-opening parabola has a maximum value at its vertex.

Now...you should have learned something about how to find the vertex of a parabola. Please continue with what I've started....find the vertex of the parabola. The x-coordinate of the vertex is the length of the rectangle which will give the maximum area, and the width is 120 - x. If you're still having trouble with this, please repost and show us your work so we can see where you might need further help.
 
juliettusa said:
A video rental service has 1000 subscribers, each of whom pays $15 per month. On the basis of survey, they believe that for each decrease of $0.25 in the montly rate, they could obtain 20 additional subscribers.At what rate will max revenue be obtained and how many subscribers will there be at that rate?
Click to expand...

Let x = the number of 25-cent reductions implemented

Now, we can write the following:

New Monthy Rate = 15 - 0.25x

New Number of Subscribers = 1000 + 20x

Revenue = (New Monthly Rate)*(New Number of Subscribers)

The revenue graph is a parabola. The vertex coordinates provide the answers to the two questions, in this exercise.
 
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