what are the dimensions of the fenced in area using the quadratic functions?

Question

A farmer marks off a rectangular area using 400m of fence. what is the greatest area she can enclose if the formula representing the swimming area is : A= 200L-L^2? What are the dimensions of the fenced in area? Give full details and steps.

Answer 1

You need to find its derivitive and set it = to zero. That gives max and mins of the Area.

So its derivitive is: dA/dL = 200-2L and that =0 for max and mins

So 200=2L and L=100

Then substitute back into formula

A= 200(100) - 100 Squared or

A=20,000-10,000 = 10,000

Then you know that Length x width = Area

So A=10,000=100 times W

and W=100

So L=100 and W=100 in meters of course.

Answer 2

The formula for the area is already given: A=200L-L^2 (which is correct, btw).

Now, you want to find a point of maximum for the function A(L). It helps (but is not absolutely necessary) to see that A(L) is an upside down parabola. Well, anyway, to find possible candidates for a point of maximum (or minimum) you have to find the "first derivative" of A and solve it for zero:

A' = 200 - 2L = 0 ==> L = 100

Now, that point IS within the boundaries for L, so it is a valid candidate. Remember, L must be more that zero but less than 200 (because the total fence length is 400).

To be sure that A(L=100) is a point of maximum (and not minimum or an inflection) the second derivative at that point must be "negative":

A'' = -2 ==> A''(L=100) = -2.

Therefore, the answer you are looking for is L=100.

Since the shape of the fenced area is a rectangle and the total fence is 400, it is a square with sides 100.

Answer 3

225 (a fifteen by utilising utilising 15 sq.) yet enable's coach that... enable one edge of the rectangle be x, and various y (imaginitive, i comprehend) 2x + 2y = 60 (for perimeter-- the entire quantity of fence) x + y = 30 y = (30 - x) section = length x width = xy = x(30 - x) this might precise be a parabola that opens downward with zeros at x = 0 and x = 30 the vertex (a optimal) may be halfway between the zeros, at x = 15 if x = 15, then y = 30 - 15 = 15 x = y = 15 ==> a sq. 15 ft on a edge, with section = 225 or, with calculus goodness: A = x(30 - x) = 30x - x^2 A' = 30 - 2x A" = -2 ==> confirming that this may be a max set A' = 0 ==> 30 - 2x = 0 ==> 30 = 2x ==> x = 15 looks huge-unfold...

Answer 4

I can not make sense of this: what is the significance of L? The maximum size of any rectangle whose perimeter is constrained is always that the two dimensions are equal, so in this case 100 meters. But what this has to do with L isn't apparent.