Thread: Algebra word problem

A framer wants to enclose a rectangular field by a fence divide it into two smaller rectangular fields by constructing another fence parallel to one side of the field.
The farmer has 3000 yards of fencing. Find the dimension of the field so that total enclosed area is a maximum. (hint let h be the height and w be the width)

then 3h+2w=3000 You want to maximize the area hw. If you solve for h in terms of w then substitute into the expression hw, you get a quadratic function (you could just as well solve for w in terms of h) Find the maximum of quadratic using one of three techniques)

bump

either take the derivative then imput the values into and see if its negative values so you have maxes...test them in and see what the absolute max is...or you can graph it and see the absolute max

Originally Posted by gumi

A framer wants to enclose a rectangular field by a fence divide it into two smaller rectangular fields by constructing another fence parallel to one side of the field.
The farmer has 3000 yards of fencing. Find the dimension of the field so that total enclosed area is a maximum. (hint let h be the height and w be the width)

then 3h+2w=3000 You want to maximize the area hw. If you solve for h in terms of w then substitute into the expression hw, you get a quadratic function (you could just as well solve for w in terms of h) Find the maximum of quadratic using one of three techniques)

Solving perimeter equation for h we get


subbing into the area equation



So we note that this is a parabola and since they are symmetric about their vertex the max in half way inbetween the w intercepts. The w intercepts are w=0 and w=1500 so the vertex is at w=750.

we could also use the vertex formula



You could also complete the square if you want.

Good luck.