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The perimeter of a rectangle is 430 meters. Express the area A as a function of the length L, and state the domain of this function

A(L) = ? m²

Answer 1

A(L) = - L² + 215L

Domain
0 < L < 215

or

(0, 215) in interval notation

Explanation:

Let L represent the length and W the width

Perimeter of a rectangle = 2(L+W) and is given as 430 meters

Therefore 2(L + W) = 430

L + W = 430/2
L + W = 215

W = 215-L

The area A is given by L x W

Substituting for W in terms of L this becomes

A = L x(215 - L)

A = 215L - L²

which can be expressed as a function of in the form
A(L) = 215L - L² or in standard form(highest degree first)

A(L) = - L² + 215L

The domain of this function are the values of L which result in a defined , real value for A

Without restrictions, the domain for L is -∞ < L < ∞. However real world facts coming into play

1. Area cannot be zero
So A(L) > 0 (upper bound on area)

-L² + 215L > 0

-L² > - 215L Move 215L to right side

L² < 215L (dividing by -1 changes signs of variables and also inequality)

L < 215 (divide by L both sides)

2. Length cannot be negative
So lower bound on L is L > 0

Together the domain is
0 < L < 215

or

(0, 215) in interval notation

Answer 2

ANSWER -

Let L be the length of the rectangle and W be the width. The perimeter is given by 2(L + W) = 430. Solving for W, we have W = (430 - 2L)/2. The area A is given by the product of the length and the width:

A(L) = L * W = L * (430 - 2L)/2 = 215L - L^2.

So the area A is a function of the length L with the equation A(L) = 215L - L^2.

The domain of this function is the set of all real numbers L such that the width W is positive, i.e., 0 < W < ∞. This implies that 0 < 215 - 2L < ∞, or 0 < 215 < 2L. Hence, the domain of the function A is L > 107.5.