Can anyone help me with these calculus problems???

For Question 1: You need to note the equations for surface area and volume. So, SA=x^2 + 4xy and V=4=x^2y. Substitute y (found from Volume relationship) into the SA equation. You need to take the derivative of SA=2x + 16x^-2. Set the derivative to zero and solve for the min/max. You find that x=2 and substituting back into volume relationship y=1. So, the dimensions of the box are 2x2x1. I would check to see if that is the min and not the max but I felt bad enough doing any easy homework question for a kid.