Optimization.?

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2–x^2 . What are the dimensions of such a rectangle with the greatest possible area?

How do I approach this?

Answer 1

Width of base = 2x (distance from -x to +x)
Height = 2-x^2

Area = u= 2x(2-x^2) = 4x - 2x^3

Maximise area:
du/dx = 4 - 6x^2 = 0
6x^2 = 4
x^2 = 2/3
x = ±sqrt(2/3)
y = 2 - x^2 = 2 - (2/3) = 4/3

Dimensions are 2sqrt(2/3) by 4/3

Answer 2

The rectangle's upper right corner could be any value from x= 0 to x=sqrt(2). Any greater x would result in y being negative and so the corners wouldn't be the upper corners any more.

The area of such a rectangle is

A=y (x-(-x)) = 2xy = 2x (2-x^2) = 4x - 2x^3.


This is itself a function and you can figure out what the maximum value is by calculus:

Change of Area with x is

dA/dx = 4 - 6x^2

You are looking for where the rate of change is zero. If the slope of this function is 0 that means it's a maximum, minimum or a saddle-point.

Where dA/dx = 0, 6x^2 = 4 x^2 = 2/3 x = 0.816 496 580 927 (and y = 4/3).

The greatest area is a square 4/3 x 4/3, or 16/9.

[Take the second derivative, d2A/dx2 = -12x. The second derivative is negative which means that the x-value where the first derivative is 0 represents a maximum.]

Answer 3

The area of a triangle is given by the length time the width. In this problem the length will be 2x and the width is 2 - x^2.

Therefore

A(x) = 2x * (2 - x^2)
A(x) = 4x - 2x^3

Take the first derivative of A(x) and set to zero to find the min or max.

A'(x) = 4 - 6x^2

0 = 4 - 6x^2
6x^2 = 4
x^2 = 2/3
x = +/- sqrt(2/3)

since the parabola is symmetrical about the y-axis

x= sqrt(2/3)

y = 2 - (sqrt(2/3))^2
y = 2 - 2/3
y = 4/3

Length = 2*sqrt(2/3)
Width = 4/3

Answer 4

Let the bottom right corner of your rectangle be x. Width of base is 2x. Height is y = 2 - x². Area is 2x(2 - x²). So

A = 4x - 2x^3
A' = 4 - 6x²

optimize by making A' = 0
6x² = 4
x² = 2/3
x = (√6)/3
base = 2(√6)/3
height = 2 - [(√6)/3]² = 2 - 6/9
height = 4/3