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Sergeeva-Olga [200]
4 years ago
6

Rectangle R has varying length l and width w but a constant perimeter of 4 ft. A. Express the area A as a function of l. What do

you know about this function? B. For what values of l and w will the area of R be greatest? Give an algebraic argument. Give a geometric arguement.

Mathematics
1 answer:
ivanzaharov [21]4 years ago
4 0
Given:
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter

Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l            (1)

Part A.
The area is
A = w*l 
   = (2 - l)*l
 A  = 2l - l²
This is a quadratic function or a parabola.

Part B.
Write the parabola in standard form.
A = -[l² - 2l]
   = -[ (l -1)² - 1]
   = -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.

The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).

Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.

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Hope this helps!
5 0
3 years ago
Read 2 more answers
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
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8x^3 + 27y^3

(2x)^3+(3y)^3

(2x+3y)(4x^2-6xy+9y^2) <----answer

7 0
3 years ago
1. One week Carlos bought 3 bags of Tabitha Tidbits and 4 bags of Figaro Flakes for $43.00. The next week he bought 3 bags of Ta
d1i1m1o1n [39]
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(43-4(5.5))/3
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Answer:

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Step-by-step explanation:

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