Area Problem.

[Alexander]

You have a linear function [math]f(x)=-\frac{2}{3}x+4[/math] where [math]0\leq x\leq 6[/math] and [math]0\leq y\leq 4[/math]. What is the maximum area of a rectangle that has one side on the line of the function?

 

I know how to optimize this, I am just having trouble finding the equation for area of such a rectangle in terms of the function.

 

I have included an image of what I am talking about. Thanks a lot.

[Asimov Pupil]

well we know that the max area for a retangle is a square right?

so now into the calculus! gimme a while

[Johnny5]

It's been a long time since I've solved a problem like this.

 

Start off with the definition of a rectangle.

 

Length times width.

 

Now, the lines which are the sides of the rectangle, have to have a slope in the XY plane, which is perpendicular to the slope of the hypotenuse, which you have drawn.

 

So the slope will be the negative reciprocal.

 

Now the slope of the given line is -2/3.

 

The negative reciprocal is 3/2.

 

Now, here is the given function:

 

[math] y = f(x) = \frac{-2x}{3} + 4[/math]

 

That is in slope intercept form, m=slope=-2/3, and y intercept is 4.

 

The x intercept is the point the line crosses the x axis, and the point where that occurs has a y coordinate of zero. So in order to find the x-intercept, you have to find the point (x,0), where 0=-2/3x+4.

 

0=-2/3x+4

2/3x=4

1/3 x =2

x=6

 

just as you have labeled.

[Asimov Pupil]

well the way i see it you can calculate an angle thaeta to find the value of the oposite side which is a side of the Square then the area is (x)^2

 

cuz according to calculus the max area is is a square