a farmer has 60m of fencing with which to construct three sides of a rectangular yard connected to an existing fence.
a. if the width of the paddock is x m and teh area inside the yard, A m^2, write down teh rule connecting A and x
b. determine teh maximum area that can be formed for the yard.

( the side of existing fence is a length)

2. Originally Posted by summna09
a farmer has 60m of fencing with which to construct three sides of a rectangular yard connected to an existing fence.
a. if the width of the paddock is x m and teh area inside the yard, A m^2, write down teh rule connecting A and x
b. determine teh maximum area that can be formed for the yard.

( the side of existing fence is a length)

Hello,

a) first make a sketch of the situation.

Since the length of the fence is 60 m the length of the rectangle is:

$l = 60 -2x$

You are supposed to know that the area of a rectangle can be calculated by

$A= l \cdot w~\implies~A(x)=(60-2x) \cdot x$

b) $A(x) = -2x^2 + 60x$ . A is a quadratic function pointing down, that means the maximum value is at it's vertex:
In general the x-value is calculated by: $x_V=-\frac{b}{2a}$ With your problem:
$x_V = -\frac{60}{2 \cdot (-2)} = 15$

Thus the rectangle has the dimension: $l = 30\ m\ \text{and } w = 15\ m$