Ms. Smith has 190 m of fencing to enclose a rectangular area.?

(a)
If width is W and length is L, then
perimeter of rectangular area = 2(L+W) = 190
(L+W) = 95
L = 95-W

Area = L*W
A(W) = (95-W)*W
A(W) = 95W - W²
function is A(W) = 95W - W²

b) Domain is (0,95)
c) Range
To find max value of the range find horizontal tangent to the function
A' = 95 - 2W
0 = 95 - 2W
2W = 95
W = 47.5
A(47.5) = 95(47.5) - (47.5)² = 2256.25

Range = (0,2256.25]

c) From b above, max area is obtained by W=47.5 m, then L = 95 - 47.5 = 47.5 m
Dimensions of 47.5 m x 47.5 m gives max area

ℓ: length of the rectangle
ω: width of the rectangle

The perimeter of the rectangle is:
p = 2.(ℓ + ω) → given that Ms. Smith has 190 m of fencing → p = 190 m
2.(ℓ + ω) = 190
ℓ + ω = 95
ℓ = 95 - ω

The surface area of the rectangle is:
s = ℓ * ω → we've just seen that: ℓ = 95 - ω
s = (95 - ω).ω
s = 95ω - ω² ← this is a function of ω where ω is the width of the rectangle

As s represents a surface area, the value is sush as:
s > 0
95ω - ω² > 0
ω.(95 - ω) > 0 → you know that ω > 0 (of course)
95 - ω > 0
- ω > - 95
ω < 95
→ ω Є ] 0 ; 95 [

Recall the surface area:
s = 95ω - ω² ← you can obtain the maximum of a function when its derivative is zero → s' = 0
s' = 95 - 2ω → then you solve for ω the equation: s' = 0
95 - 2ω = 0
2ω = 95
ω = 47.5 m

Recall: ℓ = 95 - ω
ℓ = 95 - 47.5
ℓ = 47.5 m

The surface area of the rectangle is maximum when the surface area is a square.
The maximum surface area is:
s = ℓ * ω
s = 47.5 * 47.5
s = 2256.26 m²