Question
A rectangle has its base on the x-axis and its 2 upper corners on the parabola y=12-x^2. What is the largest possible area of the rectangle?
Answers
Since this is an upside down parabola, the rectangle must be inside the parabola. The parabola and rectangle inside it are symmetrical about the y axis.
Let x be the coordinate of a corner that touches the parabola at y = 12 - x^2. The area of that parabola is
A = 2x*(12-x^2)= 24 x - 2x^3
That is a maximum when dA/dx = 0
24 -6x^2 = 0
x = +/- 2
y = 12 - 4 = 8
Maximum area = 2xy = 32
Let x be the coordinate of a corner that touches the parabola at y = 12 - x^2. The area of that parabola is
A = 2x*(12-x^2)= 24 x - 2x^3
That is a maximum when dA/dx = 0
24 -6x^2 = 0
x = +/- 2
y = 12 - 4 = 8
Maximum area = 2xy = 32