Building a rectangular room with fixed perimeter of 280ft

what dimentions would yield the maximum area?
What dimentions would yeild the minimum area?

A rectangle of largest area is a square.

So divide your perimeter by 4.

To build a room of minimum area is a rather silly question.

x = length

y = width
A = x y
x+y=140
A = x(140-x)
A = 140 x - x^2
look at shape of parabola by completing the square
x^2 -140 x = -A
x^2 - 140 x + 4900 = -A + 4900
(x-70)^2 = -A + 4900
the vertex will be at x = 70 and A = 4900 and the -A means that the parabola opens down (sheds water) so that vertex is the maximum where x = y =70 as Reiny told you.
since a negative x or y does not make much sense, the area may not get any smaller than zero when x --> 0 or y -->0 0

I apologize for my silly question. I followed your advice, revised my question and found my answer. Thank you.

Wasn't very helpful.

Yes

To find the dimensions that yield the maximum area, we can use the concept of differentiation. Let's suppose the length of the room is L and the width is W.

1. First, we need to express the perimeter in terms of L and W. The perimeter of a rectangle is calculated by adding up the lengths of all four sides. Since a rectangle has two pairs of equal sides, we can write the perimeter as:
Perimeter = 2L + 2W

Given that the perimeter is fixed at 280ft, we have:
2L + 2W = 280
Simplifying this equation, we get:
L + W = 140 (Equation 1)

2. Now we need to express the area of the rectangle in terms of L and W. The formula for the area of a rectangle is:
Area = Length × Width

So, the area of our rectangle is:
Area = L × W

3. To find the dimensions that yield the maximum area, we need to maximize the Area function. This can be done by finding the critical points where the derivative of Area with respect to either L or W is equal to zero.

Let's find the derivative of the Area function with respect to L (keeping W constant):
d(Area)/dL = W

Setting this derivative equal to zero, we get:
W = 0

Since W cannot be zero (it represents the width of the room), there is no critical point for L when W is constant.

4. Similarly, let's find the derivative of the Area function with respect to W (keeping L constant):
d(Area)/dW = L

Setting this derivative equal to zero, we get:
L = 0

Again, since L cannot be zero (it represents the length of the room), there is no critical point for W when L is constant.

5. From steps 3 and 4, there are no critical points inside the feasible region (L, W > 0).

6. Therefore, we can conclude that within the given constraints, the maximum area is achieved when both L and W are at their maximum values. From Equation 1, this occurs when L = W = 140/2 = 70ft.

Thus, the dimensions that yield the maximum area of the rectangular room are L = W = 70ft.

To find the dimensions that yield the minimum area, we can apply the same logic. Since there are no critical points within the feasible region, the minimum area will be obtained when either L or W is close to zero. However, since both L and W must be positive (L, W > 0), there is no minimum area. The smallest area is simply zero when either L or W is zero.