math challenge?

Question

A rectangle is inscribed in the triangle bounded by the x-axis and the lines whose equations are 3x+4y-24=0 and 3x-2y+12=0. Express the area of the rectangle as a function of the height and find the height for which the area is largest.

how do i do this and what does it mean by expressing as a function?

Answer 1

This is going to be a little hard to explain without being able to draw pictures, so bear with me.

You can squeeze a bunch of different rectangles into this triangle; here's how: pick some line parallel to the x axis which passes through the triangle; for instance, we could pick the line y = 3. This intersects the line 3x + 4y - 24 = 0 at the point (4, 3) and intersects the line 3x - 2y + 12 = 0 at the point (-2, 3). Now take the perpendiculars through the x axis which pass through these points -- that is, the lines x = -2 and x = 4 -- and the four lines you've drawn clearly come together to make a rectangle which is 6 x 3, so its area is 18.

Notice that, once we were given the distance 3 from the x-axis to the parallel line we chose, we were able to come up with the area of the rectangle -- in other words, the area of the rectangle was a function of the height. Now let's do this in general -- instead of the line y = 3, we'll think about the line y = h.

The line y = h (what are the upper and lower bounds on h here?) intersects the line 3x + 4y = 24 at the point

(8 - 4h/3, h)

and intersects the line 3x - 2y = -12 at the point

(-4 - 2h/3 , h)

and so forms a rectangle of height h and width

(8 - 4h/3) - (-4 -2h/3) = 12 - 2h/3

which then has area h(12 - 2h/3). This is what it means to express the area as a function of h. Of course, then you'd have to think about going through this procedure with the other two sides of the triangle. That would have been messier to type out, however.

You can demonstrate that these are all the possible rectangles by noting that, if you're inscribing a shape with four points inside a triangle, one leg of the triangle has to have two of the points on it.

So now you'd need to figure out the maximum value of each of your three functions. This is easiest when you use calculus, but I'm betting that might be a little beyond you at at the moment...

Answer 2

x1 = - 4(y - 6)/3
x2 = 2(y - 6)/3
A = y(x1 - x2)
A = y(- 4(y - 6)/3 - 2(y - 6)/3)
A = (y/3)(- 4y + 24 - 2y + 12)
A = - (1/3)(6y² - 36y)
A = - 2(y² - 6y)
A = - 2(y² - 6y + 3² - 9)
A = - 2(y - 3)² + 18
The maximum area is 6 square units when y = 3.

Answer 3

3x+4y-24=0 ----------> (24-4y)/3
3x-2y+12=0 ----------> (2y-12)/3


For any y as the height, the width is (24-4y)/3 - (2y-12)/3
=12 - 2y

So area is y(12-2y) = 12y-2y^2

Max 12y-2y^2
So 12-4y=0
y=3
Largest area = 3(12-6)=18