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earnstyle [38]
4 years ago
6

Robert has available 400 yards of fencing and wishes to enclose a rectangular area. Express the areaAof the rectangle as a funct

ion of the widthwof the rectangle. For what value ofwis the arealargest? What is the maximum area?
Mathematics
1 answer:
Ainat [17]4 years ago
6 0

Answer:

A) A = 200w - w²

B) w = 100 yards

C) Max Area = 10000 sq.yards

Step-by-step explanation:

We are told that Robert has available 400 yards of fencing.

A) we want to find the expression of the area in terms of the width "w".

Since width is "w", and perimeter is 400,if we assume that length is l, then we have;

2(l + w) = 400

Divide both sides by 2 gives;

l + w = 200

l = 200 - w

Thus, Area of rectangle can be written as;

A = w(200 - w)

A = 200w - w²

B) To find the value of w for which the area is largest, we will differentiate the expression for the area and equate to zero.

Thus;

dA/dw = 200 - 2w

Equating to zero;

200 - 2w = 0

2w = 200

w = 200/2

w = 100 yards

C) Maximum area will occur at w = 100.

Thus;

A_max = 200(100) - 100(100)

A_max = 10000 sq.yards

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<h3>Answer:  462</h3>

==================================================

Explanation:

Let's call the children A, B, C, D, E for the sake of simplicity. Also, let's have the children select in alphabetical order.

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There are 55440 different permutations possible. If order mattered, then we would stop here. However, order does not matter because the cereals aren't ranked and their position doesn't matter. All that matters is what fills the grocery cart. In other words, imagine the kids jumbling up the boxes as they push around the cart, so the order would be irrelevant.

Consider a group of 5 items. There are 5! = 5*4*3*2*1 = 120 different ways to arrange the five items in this group. So for any 5 cereals picked, there are 120 ways to order them. But again, order doesn't matter, so we have to correct for this over-counting. We do this by dividing by 120

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------------------

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We get the same answer.

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