Step-by-step explanation:
A can is a cylinder.
Volume of a can = Volume of Cylinder

Given Radius = 3.2 inches and height = 15 inches,
Volume of one olive oil can =

Volume of 5 olive oil cans = 5 x volume of one olive can

I will just leave the answer in terms of pi as I'm not sure if you need to round off your answers.

a. The gradient is


b. The gradient at point P(1, 2) is

c. The derivative of
at P in the direction of
is

It looks like

so that

Then


Since there are 2 sides to the shelf, there is 0.05*2=0.05+0.05=0.1 meters of space between the shelf and wall in total. Next, since each piece of wood is 0.3 meters wide and we have 3 of them, the total width of the wood is 0.3+0.3+<something>
Note that <width of wood>+<space between shelf and well> =width of closet
I challenge you to finish this on your own - good luck, and feel free to ask with any further questions!