A farmer wants to put a fence around his vegetable garden. Only three sides must be fenced since a rock wall will form the fourth side. If he uses 40 m of fencing, what is the maximum area possible? What dimensions should the farmer use?

Maximum area = 200m

Dimensions:

Length = 10m

Width = 20m

Such that there is only one side of 20m fenced because the other side of 20m is against the rock wall.

Step-by-step explanation:

We have the next problem.

Suppose that the vegetable area is a rectangle of length L and width W.

Such that the width side coincides with the rock wall, then we have that:

The perimeter is:

P = 2*L + 2*W

But because one of the width sides coincides with the wall, the amount of fence needed will be:

P = 2*L + W = 40m

And the area can be written as:

A = W*L

We want to find the maximum area that we can have.

First, write our two equations:

40m = 2*L + W

A = W*L

First we must isolate one of the variables in the first equation, and replace it into the second equation:

40m - 2*L = W.

then:

A = (40m - 2*L)*L = 40m*L - 2*L^2

We can write this as a quadratic equation:

A(L) = -2*L^2 + 40m*L

Now, notice that it has a leading coefficient negative, then the maximum of the area function will be at the vertex of the quadratic equation.

For a quadratic equation:

f(x) = a*x^2 + b*x + c

The vertex is at:

x = -b/2a

and the point is (-b/2a, f(-b/2a))

In this case we have:

b = 40m

a = -2

Then:

we must evaluate the area function in:

L = -40m/(-2*2) = 10m

A(L) = -2*L^2 + 40m*L

A(10m) = -2*100m^2 + 400m = 200m

Now we can find the width using one of the above equations:

W = 40m - 2*L = 40m - 2*10m = 20m

Then the dimensions that the farmer should use are:

Two sides of length 10m, and one side of length 20m

last one.

step-by-step explanation:

last one.

generally speaking, the others follow some sort of pattern that correlates to either linear (looks like a straight line) or sinusoidal (looks like an s).

the last one is just a random cloud of points that doesn't correlate to any particular known order.

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629