Answer:
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.