Question

The dimensions of a box are b, b + 1, b + 4. Find how fast the total
surface area A increases as b increases.

Answer 1

$\text{Total Surface Area of the box(A)}=2(b(b+1)+(b+1)(b+4)+(b(b+4)) \\=2((b^2+b)+(b^2+5b+4)+(b^2+4b))\\=2(3b^2+10b+4)\implies 6b^2+20b+8\\\text{Hence, } \frac{dA}{db}=12b+20$