Volume of a cuboid whose diagonal and surface area is known The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:

a) 125
b) 236
c) 361
d) 486

Solution:
$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$
How can i find volume from this information?
 A: Define 
$$\begin{cases}
p_1 = & \ell + b + h\\
p_2 = & \ell^2 + b^2 + h^2\\
p_3 = & \ell^3 + b^3 + h^3
\end{cases}
\quad\text{ and }\quad
\begin{cases}
s_1 = & \ell + b + h\\
s_2 = & \ell b + \ell h + b h\\
s_3 = & \ell b h
\end{cases}
$$
We know $p_1 = s_1 = 19$ and $p_2 = 125$.  
Apply $AM \ge GM$ to the three numbers $\ell, b, h$, we get an upper bound for $s_3$:
$$s_3 = \ell b h \le \left(\frac{\ell + b + h}{3}\right)^3 = \frac{p_1^3}{3} = \frac{19}{3}^3 \sim 254.037$$
This rules out choices (c) and (d).
Apply Cauchy Schwarz inequality to $( \sqrt{\ell}, \sqrt{b}, \sqrt{h} )$ and $( \sqrt{\ell}^3, \sqrt{b}^3, \sqrt{h}^3 )$, we get
$$p_2^2 \le p_1 p_3$$
Together with the Newton's identities


*

*$p_1 = s_1$,

*$p_2 = s_1 p_1  - 2 s_2$,

*$p_3 = s_1 p_2  - s_2 p_1 + 3s_3$


We obtain a lower bound for $s_3$:
$$s_3 = \frac13 \left( p_3 - s_1 p_2 + s_2 p_1 \right)
\ge \frac13 \left( \frac{p_2^2}{p_1} - p_1 p_2 + \frac{p_1^2 - p_2}{2} p_1 \right)
=   \frac13 \left( \frac{p_2^2}{p_1} + \frac{p_1^2 - 3 p_2}{2} p_1 \right)
=   \frac13 \left( \frac{125^2}{19} + \frac{19^2 - 3\cdot 125}{2} \cdot 19\right)
= \frac{4366}{19} \sim 229.790
$$
This rules out choice (a) and leaves us choice (b) $s_3 = 236$ as the only possible answer.
A: A hint:
The two equations $$x+y+z=19,\quad x^2+y^2+z^2=125$$
together with the conditions $x\geq0$, $y\geq0$, $z\geq0$ define an arc of a circle in ${\mathbb R}^3$. Find the minimum and the maximum of the function $f(x,y,z):=xyz$ on this arc. If exactly one of the proposed values falls between the two this has to be the solution. 
A: $l+b+h=19$
$l^2+b^2+h^2=125$
From these 2 equations, we get
$2(lb+bh+hl) = (l+b+h)^2-(l^2+b^2+h^2)
= 19^2-125 = 236 = S$
This is coincidentally the Surface area of the cuboid.
The Volume $lbh = \frac{S}{2(l^{-1}+b^{-1}+h^{-1})}$.
Assuming $l=b$, we can find $l$ and $h$ to be $\frac{19-\sqrt{7}}{3}$ and $\frac{19+2\sqrt{7}}{3}$ and the Volume is then $240.63$.
Max Volume is when $l=b=h$ and we get $254.037$, so answers $c$ and $d$ are our. All we can say is that answer $b$ ($236$) is possibly right as Volume. I don't think you can actually find the Volume. 
A: l sq b sq h sq equal to 125. because underroot of  l sq + b sq +h sq = 5 underroot 5 so by  squaring both sides we get lsq + b sq + h sq = 125.  We know that surface area of a cuboid is equal to 2(lb+bh+hl) and the sum of three sides of a cuboid is 19 so we can get the area by the formulae l+b+h whole sq - l sq + b sq + h sq that is equal to 19 sq - 125 which is equal to 236 cm sq
