Finding the dimensions of a cuboid given the volume, surface area, and diagonal

Question

Let the volume, surface area and length of the diagonal of a cuboid be as follows: $144$, $192$, $13$. Find the dimensions.

My trial: $$lbh = 144$$

$$2(lb+bh+lh) = 192 \implies lb+bh+lh = 96$$

$$l^2 + b^2 + h^2 = 13^2 = 169.$$

As $(l+b+h)^2 = l^2 + b^2 + h^2 + 2(lb+bh+lh) = 169 + 192 = 361$. So, $l+b+h = 19$.

Then?

NB: Sorry to all for posting my 1st question in a wrong manner. Thanks for helping me.

Answer 1

I am continuing from where you have left.

$$l+b+h = 19 => b+h = 19-l$$

$$lbh = 144 => bh=\dfrac{144}{l}$$

$$lb+bh+lh = 96 => l(b+h) + bh = 96$$

Now substituting the first two equations in the last equation:

$$ l(19-l) + \dfrac{144}{l} = 96$$

$$ l^3 - 19l^2 + 96 - 144 = 0$$

Solve this cubic equation to get three values of which one is length, one is breadth and the other one is height of the cuboid.

The answers are $(l,b,h) = (12,4,3)$

Answer 2

lsp's answer is a special case of Vieta's formula for cubics.

Answer 3

$l,b$ and $h$ are the roots of cubic euqation equation :

$x^3+bx^2+cx+d=0$

$lbh= \dfrac{-d}{1}$

$lb+bh+hl=\dfrac{c}{1}$

$l+b+h= \dfrac{-b}{1}$

Plug in the values and use Vieta's formula.

Aliter:

$lbh=144$

$l^2+b^2+h^2=169$

$169 \equiv 1 \mod 4$

$l^2 \equiv 0 \mod 4, b^2 \equiv 0 \mod 4 $ and $h^2 \equiv 1 \mod 4$

$ \implies$ $$l=2k$$

$$b=2q$$

$$h=2m-1$$

$lbh=144$

One of the three numbers are odd. Odd divisors of $144$ are $3$ and $1$, if $h=1$, $l+b \neq 19$ (Why?). Therefore, $h=3$.

$l^2+b^2=160$ ,$lb=48$ and $l+b=16$

$(l-b)^2=l^2+b^2-2lh= 64 \implies l-b=8$

We get $l=12$, $b=4$ and $h=3$.