Question

The parabola P: y=a-x^2, a>0, is cut by the line L: y=h, h>0, in the points A and B. the points C and D, on the x-axis, are such that ABCD is a rectangle, R.

(a)Find the coordinates for the vertices of R.
(b)Find the value(s) of h so that R has the greatest area.

Answer 1

Answer on question #34456 – Math – Analytic Geometry

The parabola P: $y = a - x^{\wedge}2$ , a&gt0, is cut by the line L: $y = h$ , h&gt0, in the points A and B. the points C and D, on the x-axis, are such that ABCD is a rectangle, R.

(a) Find the coordinates for the vertices of R.

(b) Find the value(s) of $h$ so that $R$ has the greatest area.

Solution

Let us graph this rectangular

We should find the coordinates of the points A and B. Their y-coordinates are equal to h. We can find the x-coordinates from the equation

$$h = a - x ^ {2}$$ $$x = \pm \sqrt {a - h}$$

Therefore, we get these two points: $A\left(-\sqrt{a - h};h\right)$ $B\left(\sqrt{a - h};h\right)$

Points C and D have zero y-coordinates and the same x-coordinates: $C\left(-\sqrt{a - h};0\right)$ , $B\left(\sqrt{a - h};0\right)$ .

The area of this rectangular is

$$S = 2 h \sqrt {a - h}$$

To find the maximum of this value we should find the derivative of $S$ with respect to $h$ and equate it to zero

$$S ^ {\prime} = 2 \sqrt {a - h} - \frac {h}{\sqrt {a - h}} = \frac {2 a - 2 h - h}{\sqrt {a - h}} = \frac {(2 a - 3 h)}{\sqrt {a - h}} = 0$$ $$2 a - 3 h = 0, \quad \Rightarrow h = \frac {2 a}{3}.$$

At this point the function $S(x)$ has the maximum.

Answer: (a) $A\left(-\sqrt{a - h};h\right)$ , $B\left(\sqrt{a - h};h\right)$ , $C\left(-\sqrt{a - h};0\right)$ , $B\left(\sqrt{a - h};0\right)$ .

(b) $h = \frac{2a}{3}$

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