The total length of all $12$ sides of a rectangular box is $60$.
The surface area of the box is given to be $56$. Find
$(i)$ the length of the longest
diagonal of the box
$(ii)$ the volume of the box
Let the one of the sides be $a$ and the other be $b$
$$8a+4b=60 $$ and $$2a^2+4ab=56 $$
Solving theses I get values of $a=2/3$ and $b=16/3$ but they do not match with the answer given for longest diagonal$=13$
It is also given in the answer that the volume cannot be determined. But having got $a$ and $b$ we should be able to find the volume.
It seems to me that there are three dimensions, not two. Call the side lengths $\{a,b,c\}$. We have $$a+b+c=15\;\;\&\;\;ab+ac+bc=28$$
The longest diagonal is $d$ and we easily see $d^2=a^2+b^2+c^2$. But then we compute: $$225=(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=d^2+56$$
Thus we can compute $d$.
I see no way to compute the volume...we have two equations in three unknown and the three lengths are not determined by the given data.
To be complete: to see that the volume is not determined from the given data, we'll construct two boxes with different volumes that meet the constraints. For the first we'll assume, as the OP does, that $a=c$. That gives us $a=\frac 13\times \left(15-\sqrt {141}\right)=c,\;b=15-2a$. (Note: this differs from the values given by the OP). We quickly get volume $\sim 14.02$ in this case. As an alternative, let us suppose that $c=2a$. Working numerically, we get $\{a,b,c\}=\{.69801,12.91,1.396\}$ and in this case the volume is $\sim 12.58$.