Rectangle in the parabola and line

Geometry Level 4

What's the maximum rectangle area that is covered with parabola $y^2=4x$ and line $x=16$ ?

The maximum rectangle area can be expressed as $\dfrac{a\sqrt{b}}{c}$ , where $a$ , $b$ , $c$ are positive coprime integers.

Find the value of $a+b+c$ .

The answer is 524.

1 solution

David Vreken
Mar 30, 2021

Consider the following rectangle:

The top left corner is on the parabola $y^2 = 4x$ , so let its coordinates be $(k, 2\sqrt{k})$ .

Then the base of the rectangle is $16 - k$ and the height of the rectangle is $4\sqrt{k}$ , which makes its area $A = 4(16 - k)\sqrt{k}$ .

The maximum area is when $A' = \cfrac{2(16 - 3k)}{\sqrt{k}} = 0$ , which is when $k = \frac{16}{3}$ . At this $k$ value, the area is $A = 4(16 - \frac{16}{3})\sqrt{\frac{16}{3}} = \frac{512\sqrt{3}}{9}$ .

Therefore, $a = 512$ , $b = 3$ , $c = 9$ , and $a + b + c = \boxed{524}$ .

I might be missing something, but is it obvious that the rectangle must have two sides parallel to the axis of symmetry of the parabola ? Also is it necessary that all 4 vertices must either lie on the line (x=16) or the parabola ?

Sathvik Acharya - 2 months, 2 weeks ago

I must confess that I assumed that this must be the case for a maximum area rectangle in the parabola, but don't have a proof for it yet.

David Vreken - 2 months, 2 weeks ago

Rectangle in the parabola and line

The answer is 524.

1 solution

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