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A rectangle has area 25 m2. express the perimeter of the rectangle as a function of the length l of one of its sides. p(l) = state the domain of p. (assume the length of the rectangle is larger than its width. enter your answer using interval notation.)

Respuesta :

sqdancefan
a) The area of a rectangle is the product of length and width:
   A = l·w

This formula gives us a way to represent the width in terms of area and length. Divide by l to get
   A/l = w

Now we know that the perimeter of a rectangle is the sum of the lengths of its sides.
   P = l + w + l + w
   P = 2(l + w)
We want to find a formula that just has l on the right, so we can use the expression for w from above.
   P = 2(l +A/l)

Substituting the given number for A, the function p(l) can be written as
   p(l) = 2(l +25/l)
   p(l) = 2l +50/l


b) For length greater than width, we have
   l > w
   l > A/l
   l² > A
   l > √A
   l > √25

This means the domain of p(l) is
   l ∈ (5, ∞) . . . . . meters
facundo3141592

For a general rectangle of length L and width W, the perimeter is given by:

[tex]P = 2L + 2W[/tex]

And the area is given by:

[tex]A = L\cdot W[/tex]

With these properties, we will found that the domain of the perimeter as a function of L is (5m, ∞)

Here we know that:

The area is 25m^2

The length is larger than the width, so we have L > W.

Because both L and W are the measures of a rectangle, we also should define:

L > 0m

W > 0m.

Now we want to find the domain of the function P(L), the perimeter as the length of the rectangle, where we need to remember that the domain is the set of the possible inputs of a function.

From the equation of the area, we can get:

[tex]25m^2 = A = L\cdot W\\\\25m^2 = L\cdot W\\\\25m^2/L = W[/tex]

Now we can replace that width in the equation for the perimeter:

[tex]P(L) = 2W + 2L = 2(25m^2/L) + 2L[/tex]

Now let's find the possible values of L.

Because L > W, we can find a lower bound of L if we assume:

L = W

Then the area is:

[tex]25m^2 = L^2 \\\\\sqrt{25m^2} = L = 5m[/tex]

This means that we must have:

L > 5m.

Now we need to find an upper bound.

Notice that we have:

W > 0m

But W can get really close to 0m, so if we take:

[tex]25m^2 = L \cdot W\\\\L = 25m^2/W\\ \\\lim_{W \to \ 0m} L = \lim_{W \to \ 0m} 25m^2/W = \infty[/tex]

This means that we do not have an upper bound, as L can be really large if W is enough small.

Then the domain of P(L) will be (5m, ∞)

If you want to learn more, you can read:

https://brainly.com/question/16875632

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