A trapezoid is drawn inside a semi circle cross section with the upper base length, being the length of the circle diameter, $d$, and a lower base, $x$, touching the lower sides of the semi circle.
What is the maximum length of the lower base, $x$, of the trapezoid if its area is to be a maximum?
What is the maximum are of the trapezoidal cross section in terms of $d$?
So far I have determined from $A=(\frac{x+d}2)h$, that $h$ is equal to $\sqrt{(\frac{d}2)^2 – (\frac{x}2)^2}$
And therefore $A$ in terms of $x$ & $d$.
$$A=(\frac{x+d}2)(\sqrt{(\frac{d}2)^2 – (\frac{x}2)^2}$$
$$A=(\frac{x+d}2)(\sqrt{\frac{d^2}4 – \frac{x^2}4})$$
$$A=(\frac{x+d}2)(\frac{\sqrt{d^2 – x^2}}2)$$
$$A= \frac14 (x+d) (\sqrt{d^2 – x^2})$$
this is where I got stuck… help?
Best Answer
Answer:
I really do not know what the problem is when you got the steps absolutely right.
If differentiation is the problem:
$$\frac{dA}{dx} = \sqrt{d^2-x^2} + (x+d) \frac{-2x}{2\sqrt{d^2-x^2}}$$
$$\frac{dA}{dx} = 0$$
Assuming two functions $U = x+d$ and $V = \sqrt{d^2-x^2}$
$$\frac{dA}{dx} = V\frac{dU}{dx}+U\frac{dV}{dx}$$ $\frac{dU}{dx} = 1$ and $\frac{dV}{dx} = \frac{-2x}{2\sqrt{d^2-x^2}}$
$$\frac{dA}{dx} = \sqrt{d^2-x^2}+(x+d)\frac{-2x}{2\sqrt{d^2-x^2}} = d^2-x^2 -x(x+d) $$ $$= d^2 - 2x^2-xd = 2x^2 +xd-d^2=0$$ This a quadratic with x as the variable and two roots of x in terms of d is what follows.
the two roots of a quadratic$ax^2+bx+c = 0$ is as follows:
$$ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$.
In our case $a = 2$, $b = d$ and $c = -d^2$
\begin{align} x &= \frac{-d\pm\sqrt{d^2 + 8d^2}}{4}\\ &= \frac{-d\pm\sqrt{9d^2}}{4}\\ &= \frac{-d\pm3d}{4}\\ &= \frac{d}{2} \text{ or } \frac{-2d}{2}\text{ (rej.)}\\ \end{align}.
Now substitute the value of d in the area and find A in terms of d.
Goodluck
Thanks Satish