Calculus Help!!

Question

A farmer wants to fence in a rectangular field. The farmer must use a different type of fencing for each pair of parallel sides of the field. One type costs $10 per metre, and the other type costs $5 per metre. Assuming the farmer can spend $800 on fencing in total, what are the dimensions of the largest possible field she could make?

Answer 1

Hi Josh,

Let's understand what the question is asking here. It says that we would like to create a fence around a rectangular field. If different types of fencing is used for each pair of parallel sides, this means that we're paying $10 for one pair of opposite sides (length) and $5 for the other pair of opposite sides (width).

Let x be the number of meters we purchase for the lengths and y be the number of meters we purchase for the widths. We purchase $800 total when summed together, so this can be expressed by the following equation then:

10x + 5y = 800

Next, the area of a rectangle will be length (x/2) times width (y/2) - remember that x and y are each the total number of meters from both opposite sides (all 4 sides), so we need to divide by 2 from each to get the length and width of one of the sides - so we can use this as our other equation since we're going to want to know how to maximize our area at a certain point:

A = (x/2)*(y/2)

Now, if we understood what y was, we can substitute that for y in A = (x/2)(y/2) so that we're evaluating for one variable first!

10x + 5y = 800 ---> Solve for y.

5y = -10x + 800

y = -2x + 160

Now we can substitute as noted above!

A = (x/2)*(y/2) = (xy)/4

A = (x(-2x + 160))/4

A = (-2x^2 + 160x)/4

A = (-1/2)x^2 + 40x

Notice how we have a quadratic equation now. Because of the negative leading coefficient, we know that the parabola is facing downward. This means our vertex is going to be the point that gives us our maximum! At the vertex, the derivative (instantaneous slope) is 0 because you're at the top of the curve. Therefore, we can now find the derivative of this equation and set it equal to 0.

A = -(1/2)x^2 + 40x ---> Use the Power Rule of derivatives

A' = -x + 40 ----> Derivative equation

0 = -x + 40 -----> Set derivative equation equal to 0 and solve for x first.

x = 40

If x (length of both sides combined) = 40 meters, and we purchased $800 total worth of fencing, then plugging this back into 10x + 5y = 800, we have:

10(40) + 5y = 800

400 + 5y = 800

5y = 400

y = 80

However, remember that this is the total of both lengths (opposite sides) and widths (opposite sides), so we need to take each and divide it by 2.

40/2 = 20 and 80/2 = 40, so the dimensions of the largest rectangular fence she can make around the field is 20 meters x 40 meters, which gives you a maximum area of 800 sq. meters.

Any questions, let me know.

Thanks!

Answer 2

800/5= 160 meters of fence within the budget

160/4 = 40 m by 40 m gives maximum area = 1600 m^2

Unless you have to use $10 fencing for 2 sides then

a different answer

800 some for $10 fencing some for $5 fencing

40 m for chapter, 20 for double that price

40x20 = 800 max area

Answer 3

20a+10b=800

a=80-b

area=80b-b²

d(area)/db=80-2b

set derivative to zero

,,, area is maximum when b=40 and a=20