A = WL
2W + 3L = 1200
W = 600 - 3L/2
A(L) = (600 - 3L/2)L = 600L - 3L^{2}/2
take the derivative of A with respect to L (dA/dL) and set it equal to 0, then solve for L.
Hi. I thing I'm missing something simple, but can anyone help me solve the following:
A role of fence wire 1200m long is to be used to enclose and seperate two identical adjoining rectangular paddocks (which share one side-width).
Find the dimensions of the enclosure, so the total area is maximised.
Thanks in advance for the help.
Hello, elmidge!
Did you make a sketch?
A roll of fence wire 1200m long is to be used to enclose and seperate
two identical adjoining rectangular paddocks (which share one side-width).
Find the dimensions of the enclosure so the total area is maximised.
Let = total length of the paddocks.
Let = width of the paddocks.
We have: . .[1]Code:: - - - - - x - - - - - : *-----------*-----------* | | | | | | y| |y |y | | | | | | *-----------*-----------* : - - - - - x - - - - - :
The area is: .
Substitute [1]: .
This is a down-opening parabola:
Its maximum is at its vertex: .
Hence: .
Substitute into [1]: .