What's the maximum area you can get for a rectangle with two sides along the x and y axes, and the opposite vertex in the first quadrant along the line y = 20 – 4x?

Remember that a triangle rectangle of length L and width W has an area:

A = W*L

In our rectangle, we have two sides along the x and y axes.

So one of the vertices of our triangle rectangle is the point (0, 0)

And the other vertex, is along the line:

y = -4x + 20

So, if the opposite vertex is at the point:

(x₁, y₁)

We can define the length as the difference between the x-values of each vertex.

L = (x₁ - 0) = x₁

And the width, similarly, as:

W = (y₁ - 0) = y₁

Such that the point (x₁, y₁) is a solution for the equation y = -4x + 20, then we have:

y₁ = -4x₁ + 20

Then we can rewrite the width as:

W = -4x₁ + 20

Now, we can write the area of our rectangle as:

A = (x₁)*(-4x₁ + 20)

A = -4*x₁^2 + 20*x₁

Now we want to maximize the area, notice that the area is given by a quadratic equation with a negative leading coefficient.

Thus, the maximum will be at the vertex of that quadratic equation.

Remember that for a general quadratic equation:

y = a*x^2 + b*x + c

The x-value of the vertex is:

x = -b/(2*a)

so, in our case, the x-value of the vertex will be:

x₁ = -20/(-4*2) = 20/8 = 5/2

Now we can evaluate this in our area equation:

A = -4*(5/2)^2 + 20*(5/2)  = 49.36

This is the maximum area of the rectangle.

6 meters

step-by-step explanation: