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 oblong, no longer as a quadrilateral. a million) Width = x, length = (40 - 2x)/2 = 20 - x area, A = x(20-x) = 20x - x^2 For maxima/minima, differentiating with comprehend to x and placing it to 0: dA/dx = 20-2x = 0 x = 10 = width length = (40 - 2x)/2 = 10 Max area = 10 x 10 = one hundred 2) For maximumu area, length = width = x = 10 improve the width via h Width = 10 + h length = (40 - 2(10+h))/2 = 10 - h area = (10 + h)(10 - h) = one hundred - h^2 further if one area is decreased via h, area = one hundred - h^2 If one area is bigger or decreased via h, the fenced-in area would be decreased via h^2. 3) A = x(20 - x), the place x = width 2nd area: Width = x length = (40 - 2x), as we are watching for in straightforward terms 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 = 2 hundred

Answer 3

How about instead of the solution, I'll show you how to do it eh?

First off you need to set up a system of eqns:

let x = length and y = width

2x+2y=40
xy=Area

Now express x in the second eqn in terms of y, by solving the first eqn:

x=20-y

Now plug it into the 2nd eqn:

(20-y)(y)=Area = 20y-y^2

This is the equation for all possible pens. Now just find the maximum of this quadratic (either by graphing or vertex).


The bonus is similar but instead you have:

2x+y=40
xy=area

Simply do the same as the first question; expressing x in terms of y, distribute the multiplication, and then finding the maximum.