Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the line 𝑓(𝑥)=24−3𝑥 and sides parallel to the axes.

Question

Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the line f(x)=24-3x , and sides parallel to the axes. 

The maximum possible area is

Answer 1

This is an optimization problem.

First, realize that the length (x) and height (y) of the rectangle is essentially the x,y point on the curve, because the distance to the origin is simply the value of those coordinates.

For example, if the point is 3,7 the length would be 3, the height 7, and the area is 3*7 = 21.

Lets start with constraints:

Area will be

A = L*W = x*y (y in this case is f(x) )

we plug f(x) in for y and get our area equation in terms of just x

A(x) = x*f(x) = 24x - 3x^2

To find the max, we need to take the derivative and set it = 0 to find the critical points.

A '(x) = 24 - 6x

0 = 24 - 6x

x = 4 (this is the value where either the max or min will occur, but in this case there is only one, a max)

knowing x is the ideal length, we can find y as the ideal height y by plugging our x into f(x)

y = f(x) = 24 - 3(4) = 12

so the max area is 4*12 = 48

I recommend graphing the original area function to see how the curve reaches a max at that point of x = 4. try literally googling "graph 24x - 3x^2" Please let me know if you have other questions!