Maximizing the Area of a Rectangle Under Two Functions

[riesenman]

Hi, I've gotten stuck on this problem and was hoping you could help me.

Let 0 < X₁ < X₂ be points on the X-axis. Consider the rectangle ABCD with vertices C and D on the lines given by "y = -x + 15" and "y = 4x + 5" respectively. C and D are the points of intersection with the given lines. Find the maximum area of such a rectangle.

Here's a graph showing the two functions and relevant points:

I figured that you can get the point of A by plugging it into the function (4x+5):
A = 4a+5
and then you can express B as A by setting y=-x+15 equal to 4a+5 like so:
4a + 5 = -(b) + 15
4a - 10 = -(b)
B = -4a+10

Then to get the width of AB you subtract the two values and get
(b - a)
-4a+10 - (4a+5)
Width = -8a - 15

However, I'm not sure where to go from here.

 

[Ishuda]

Hi, I've gotten stuck on this problem and was hoping you could help me.

Here's a graph showing the two functions and relevant points:
View attachment 5298
I figured that you can get the point of A by plugging it into the function (4x+5):
A = 4a+5
and then you can express B as A by setting y=-x+15 equal to 4a+5 like so:
4a + 5 = -(b) + 15
4a - 10 = -(b)
B = -4a+10

Then to get the width of AB you subtract the two values and get
(b - a)
-4a+10 - (4a+5)
Width = -8a - 15

However, I'm not sure where to go from here.

Let me walk through the problem to catch up with you: Point a is (x₁, 0), Point b is (x₂, 0), Point c is (x₁, y₁), and Point d is (x₂, y₂). You said x₁ was a and solved for y₁ (which you called A) as the intersection of the lines x=x₁ and y=4x+5. Since y₂ must equal y₁, you can find x₂ (which you call b) in terms of x₁ by solving
y₂=-x₂+15=y₁=4x₁+5
and you got x₂=10-4x₁. You now have the area A as
A = width * height = (10 - 5 x₁) * height
and this is where you are stuck (and you got the width wrong).

Well the height is just y₁, so we have
A(x₁) = (10 - 5 x₁) * (4 x₁ + 5) = -20 x₁² + 15 x₁ + 50
and we want to maximize A with respect to x₁.

Can you go from there?