A bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open.
Find its dimensions in order that the least amount of iron sheet may be used in its construction.
My approach:
Lets say all dimensions are each $ > 0$;
Volume of the bucket - $V=\frac{lwh}{2}$ ,where l-length of toblerone, w-width, h-height;
Total exterior area of the bucket - $A= 2\frac {hw}{2}+2l(\frac{w^2}{4}+h^2)^\frac{1}{2} \Rightarrow$ $A= hw+\frac{4V}{hw}(\frac{w^2}{4}+h^2)^\frac{1}{2} $;
$\frac{\partial A}{\partial h} = w - \frac {4V}{h^2w}(\frac{w^2}{4}+h^2)^\frac{1}{2}+\frac {4V}{w(\frac{w^2}{4}+h^2)^\frac{1}{2}}$;
$\frac{\partial A}{\partial w} = h - \frac {4V}{hw^2}(\frac{w^2}{4}+h^2)^\frac{1}{2}+\frac {V}{h(\frac{w^2}{4}+h^2)^\frac{1}{2}}$;
Is this the right path to continue? Looks too messy in my eyes.
It is not so messy. Use $w=\lambda h$ to get $$\frac{\partial A}{\partial h} =h \lambda -\frac{2 \lambda V}{h^2 \sqrt{\lambda ^2+4}}=0\implies h -\frac{2 V}{h^2 \sqrt{\lambda ^2+4}}=0\tag 1$$ $$\frac{\partial A}{\partial w} =h-\frac{8 V}{h^2 \lambda ^2 \sqrt{\lambda ^2+4}}=\tag 20$$ Subtract one from the other to get $$\frac{2 \left(\lambda ^2-4\right) V}{h^2 \lambda ^2 \sqrt{\lambda ^2+4}}=0\implies \lambda=2$$ Use this in $(1)$ to get $h$ and once done reuse $w=\lambda h$.