calculus question?

Question

Jeff is making a toy box for his nephew. He will use pine for the square base and sides, but oak for the top. Pine costs $16/m2 and oak costs $20/m2. If the volume of the toy box is to be 1.3 m3, find the amount of oak that is needed if the cost of all the wood for the toy box is to be as low as possible.

Answer 1

I don't see any calculus involved in your question. A box that is square on it's base and on all of its sides, will be cubic. The oak top will be the same size as each of the pine sides. Since each edge of this cube must be 1.3 meters, then the oak top must be 1.3 m2 in area regardless of cost concerns. I'm assuming that we are disregarding the thickness of the woods.

If the adjective, square, is not intended to modify the noun, sides, then you have a calculus problem and need someone else to help you. d:c)

Answer 2

Do you mean that only the base is square, while the sides are not, In that case two equations can be set up :
A = (p+q).a^2 + p. 4ab ......(1)
V = b.a^2 ......(2)
where, a is the length/width of the base, b the height, p & q are the costs per unit area of pine and oak respectively, A is the area of wood used and V is the volume of the box.
Eliminating 'b' between (1) & (2) and simplifying,
A = 4pV/a + (p+q).a^2 .......(3)
For extremum 'A', put dA/da = 0
-4pV/a^2 + 2(p+q).a =0, which simplifies to
a^3 = 2pV/(p+q) .......(4)
Now, d^2V/da^2 = 8pV/a^3 + 2(p+q) .......(5)
On plugging the value of a^3 from (4) into (5), we get, after simplifying,
d^2V/da^2 = 6(p+q) .......(5)
Since this is positive, (4) corresponds to the minimum in 'A'.
From (4), a = [2pV/(p+q)]^(1/3) .....(6)
Plugging this into (2) ,
b = V/[2pV/(p+q)]^(2/3) ......(7)
The area of oak needed for the minimum total cost is then
A' = a^2 + 4ab
I hope you will be able to calculate this area by substituting the given values of p, q and V.

Answer 3

With the assumption that the box has the shape of a cube, 1.19 m^2 of oak is needed.
1.3^(1/3)=1.09
1.09^2= 1.19

Answer 4

Are the sides squares as well?