So I have this problem that I'm trying to solve:
A 144-cm wire is cut into 8 pieces and welded together to form a pyramid with $b = 1.9a$ in the rectangular base. Determine the dimensions a and b such that the pyramid’s volume is maximized. Define a=4:0.01:14 and use this vector for calculating volume $V = abh/3$ (after expressing h in terms of a). 3 Use the function max to determine the greatest volume.
Assuming it's a right pyramid, I know how to solve for everything but $h(a)$. So I've tried to use the system of equations:
$144 = 2a + 2b + 4s$, given that s is one of the edges from the base to the point.
$b = 1.9a$
$s = \sqrt{h^2+r^2}$, given that r is the radius from the center of the base to a corner.
$r = \sqrt{(a/2)^2+(b/2)^2}$
which led me to:
$ h = 0.229416\sqrt{18.05a^2 - 1983.6a + 24624}$
I'm very confused though, because I solved it about three weeks ago the first time and calculated $h=3a$, which is much simpler and closer to what I would expect.
$h=3a$ cannot be right as $h$ must be a decreasing function of $a$ over the range of interest
$$144=2a+2b+4s\\144=2a+3.8a+4\sqrt{\frac 14a^2+\frac{3.61}4a^2+h^2}\\ \frac 14(144-5.8a)^2=4.61a^2+4h^2\\h=\frac 12\sqrt{\frac 14(144-5.8a)^2-4.61a^2}\\ h=\frac12 \sqrt{3.8 a^2 - 417.6 a + 5184}$$