Write a function to represent the area as the function of the width...?

Question

Situation: Owners of a farm want to fence in an area for goats. They want to use up 40 meters of fencing they purchase.

1) What id the maximum area they can provide for the animals with 40 meters of fence if they want the pen to be a quadrilateral?

2) How does changing the width affect the area?

3) Write a function to represent the area as the function of the width?



Bonus: Now imagine that the fenced-in area is adjacent to a building. You only need to fence three sides.

Solution would be highly appreciated.

Answer 1

In order to answer this question, it's best to do #3 first.

If the farmers have only 40m of fence, then the perimeter of the area will be 40m. The formula for the perimeter of a rectangle is p = 2l + 2w, so 40 = 2l + 2w, or dividing both sides by 2, 20 = l + w. You can also solve this for l: l = 20 - w.

Then since the area is length times width, A = lw, you can substitute (20 - w) from the first equation to get a formula for A in terms of w only:

A(w) = (20 - w)w = 20w - w²

Now, if you're familiar with parabolas, this is a parabola. The vertex (h, k) is the maximum value of this parabola:

y = (x - h)² + k

To put A(w) in this format, complete the square:

A(w) = -w² + 20w = -(w² - 20w)
= -(w² - 20w + 100) + 100
= -(w - 10)² + 100

So the vertex is (10, 100). This means that the maximum area occurs when x = 10 and that maximum area would be 100.

Also, looking at the equation in the parabola form, you should be able to see that any value of w other than 10 will reduce the value of the area.

So the answers again are:

1) 100 square meters.
2) When the width is between 0 and 10 meters, increasing the width increases the area, to a maximum of 100 m² when the width is 10m. Between 10 and 20 meters, increasing the width decreases the area.
3) A(w) = 20w - w²

If the fenced-in area is adjacent to a building, the perimeter equation would be p = w + 2l, so 40 = w + 2l, so l = 20 - 1/2 w.

Then A(w) = (20 - 1/2w)w = 20w - 1/2w².

Finding the vertex for this parabola:

A(w) = -1/2(w² - 40w)
= -1/2(w² - 40w) - 200 + 200
= -1/2(w² - 40w + 400) + 200
= -1/2(w - 20)² + 200

So the vertex here is (20, 200); therefore the area is at its maximum (200 m²) when the width is 20.

Answer 2

I used the fence as rectangular, not as a quadrilateral.
1) Width = x, length = (40 - 2x)/2 = 20 - x
Area, A = x(20-x) = 20x - x^2
For maxima/minima, differentiating with respect to x and setting it to zero:
dA/dx = 20-2x = 0
x = 10 = width

length = (40 - 2x)/2 = 10

Max area = 10 x 10 = 100

2) For maximumu area, length = width = x = 10
Increase the width by h
Width = 10 + h
Length = (40 - 2(10+h))/2 = 10 - h
Area
= (10 + h)(10 - h)
= 100 - h^2
Similarly if one side is decreased by h, area = 100 - h^2

If one side is increased or decreased by h, the fenced-in area will be decreased by h^2.

3) A = x(20 - x), where x = width

Second part:
Width = x
Length = (40 - 2x), as we assume only one length
Area A = x(40 - 2x) = 40x - 2x^2
dA/dx = 40 - 4x = 0, for maxima/minima
x = 10
Length = 40 - 2 . 10 = 20
Max area = 10 . 20 = 200

Answer 3

How approximately as a replace of the answer, i will instruct you the thank you to do it eh? first of all you're able to set up a gadget of eqns: permit x = length and y = width 2x+2y=40 xy=section Now exhibit x in the 2nd eqn in terms of y, via fixing the 1st eqn: x=20-y Now plug it into the 2nd eqn: (20-y)(y)=section = 20y-y^2 it somewhat is the equation for all obtainable pens. Now basically locate the optimal of this quadratic (the two via graphing or vertex). The bonus is comparable yet as a replace you have: 2x+y=40 xy=section basically do a similar because of the fact the 1st question; expressing x in terms of y, distribute the multiplication, and then looking the optimal.

Answer 4

1)max area would be 100m, in a 10 by 10 perfect square
2)Because multiplication is different than adding, obviously, therefore the width will affect the output of the area, even though the output of the width doesn't.
3) (w)=A/ l ...... width equals Area divided by length

Bonus: Area=200 ...l=10 , w=20

Answer 5

might get you started:

1) 40 = 2w +2l

l= length w= width

graph this see the maximum

2) the wider it is the bigger the area...?

3)w = 20 - l


bonus: 40 = w +2l