?

[rachael724]

Yesterday I posted a problem similar to this one. I got really confused with the explanation. My teacher gave me this one to do which is similiar. I am somehow getting myslfe more confused after reading back on yesterdays verbage. Can someone help clear things up? THanks.

A rancher has 260 feet of fencing. Using this for three sides of a rectangular corral where the fourth side will be a river with a straight shore, what is the maximum number of square feet of corral space that she can expect?

 

[stapel]

The only difference between this exercise and yesterday's is that you're using "260" instead of "192". Start with the complete worked solution from yesterday. Note that the maximum will be the vertex of the quadratic "area" equation you were given. Plug in "192" instead of "260".

Where are you stuck?

Eliz.

 

[rachael724]

I'm sorry I'm so stupid is the answer 5832 ?

 

[Guest]

ok --start it from the beginning (again)
Area = x . y where x and y are the 2 sides

then 260 = 2x + y , this is your total fencing length.

then rewrite this as y = 260 - 2x

Also the area enclosed is given by the following,
A = x . y

sub in the y= 260 -2x into the Area equation.

A = x (260 - 2x)
A= 260x - 2x^2

You need to now look at the dA/dx for this and let it equal 0 to find the turning points(max and min's). Solve for x, then use this to find y, then you have the Area (which is not the answer you listed)

Back to you

 

[Denis]

Looking at your previous one; did you understand this:
dA/dx = 192 - 4x ?

Are you in a calculus class?