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 parabola f(x)=1029−7x2, and sides parallel to the axes.

Question

What is the maximum possible area?

Answer 1

So Carsyn W.,

Don't just go guessing! Sketch a possible rectangle; write expressions for the width and height (length) of that rectangle. You know exactly where the far-corner point is, by cooridinates (they are in terms of x, aren't they). So write the expression for the area (product of width and height functions), differentiate it once by d/dx, and solve for the "zero". That will be a local maximum, won't it. That gives you a value for x; the original function gives you the corresponding value for y.

Piece of cake -- if you cut your cake in rectangles? If you don't, but remembered to grease and flour the pan, it could still "pan out"?

--Cheers, --mr. d.

Answer 2

f(x) = 1029 - 7x²

To find the x-intercepts solve 1029 - 7x² = 0 and x = 12.1 (quadrant 1)

The vertex of this parabola which opens down is at (0,1029).

So, on a rectangular coordinate system draw the right branch of the parabola intersecting the x-axis at 12.1

Now draw a possible rectangle where the diagonal from the origin intersects the parabola in quadrant 1.

The area of this rectangle is length(x) times width (y=1029 - 7x²)

So, A(x) = x(1029 - 7x²) or 1029x - 7x³

To maximize the area find the derivative, A¹(x) = 1029 - 21x²

Now find the critical number by setting the derivative = 0 and solve.

1029 - 21x² = 0

21x² = 1029

x² = 49 and x = 7, -7 (reject the -7 as we are in quadrant 1)

So we now know that the area is maximized at x = 7.

Substituting into the area formula above, A = 1029(7) - 7(7³) = 4802 units²