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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?

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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.

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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.

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$l+b+h=19$ $l^2+b^2+h^2=125$ 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.

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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

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  • $\begingroup$ try to format a little bit better, how you phrased the problem is unclear.. $\endgroup$
    – Franco
    Mar 15 '15 at 5:29

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