Please help with this problem

G

Guest

Guest
I need help with this problem.. can someone please walk me through it?

You have 100 meters of fencing material to enclose a rectangular plot. Find the largest possible area which can be enclosed by the fencing. Use all three methods- graphical, numerical, and algerbraic/
 
Hello, blinded_by_beauty!

You have 100 meters of fencing material to enclose a rectangular plot.
Find the largest possible area which can be enclosed by the fencing.
Use all three methods- graphical, numerical, and algerbraic/
Click to expand...
It's probably that last sentence that's throwing us off.
I'm not familiar with "numerical" ... I can guess what the others are.
Code: 
        x
 * - - - - - - *
 |             |
 |             | 50 - x
 |             |
 * - - - - - - *
        x
Let x = length of the plot (in meters).

Then the two lengths use up 2x meters of the fencing,
. . . leaving 100 - 2x meters for the two widths.

This means each width is (100 - 2x)/2 = 50 - x meters.

The area of a rectangle is: Length x Width
. . . so we have: . A .= .x(50 - x)

We have a <u>parabola</u>: . A .= .- x<sup>2</sup> + 50x</sup>

We know that this parabola opens down (don't we?)
. . . so its vertex must be the highest point.

We're expected to know that the vertex is at: . x = -b/2a
For this problem, a = -1, b = 50, so we have: . x = -50/(2(-1) = 25

We have maximum A when x = 25.
This leaves 50 - x = 25 for the width.

Therefore, for maximum area, make the plot a 25-by-25 square.
 
Hi, this is an add on to the previous question, would somebody mind helping?

What is the maximum area you could enclose if the plot could be any shape?
 
The max area would be a circle with a circumf = 100 m
 
You must log in or register to reply here.
Top