GT
1 answer:
You might be interested in
Answer:
½
Step-by-step explanation:
Draw a picture of the triangle with the rectangle inside it.
Let's say the width and height of the triangle are w and h (these are constants).
Let's say the width and height of the rectangle are x and y (these are variables).
The area of the triangle is ½ wh.
The area of the rectangle is xy.
Using similar triangles, we can say:
(h − y) / h = x / w
x = (w/h) (h − y)
So the rectangle's area in terms of only y is:
A = (w/h) (h − y) y
A = (w/h) (hy − y²)
We want to maximize this, so find dA/dy and set to 0:
dA/dy = (w/h) (h − 2y)
0 = (w/h) (h − 2y)
0 = h − 2y
y = h/2
So the width of the rectangle is:
x = (w/h) (h − y)
x = (w/h) (h − h/2)
x = (w/h) (h/2)
x = w/2
That means the area of the rectangle is:
A = xy
A = ¼ wh
The ratio between the rectangle's area and the triangle's area is:
(¼ wh) / (½ wh)
½
So no matter what the dimensions of the triangle are, the maximum rectangle will always be ½ its area.
