An open rectangular box having a volume of is to be constructed from a tin sheet.Find the dimensions of such a box if the amount of material used in its construction is to be minimal. Hint: Let the dimensions of the box be x by y by z.Then, xyz = 108 and the amount of material used is given by S = xy + 2yz + 2xz.Show that Minimize f(x, y)
A)x = 1; y = 1; z = 2
B)x = 2; y = 2; z = 1
C)x = 2; y = 1; z = 1
D)x = 2; y = 1; z = 2

Question

An open rectangular box having a volume of  is to be constructed from a tin sheet.Find the dimensions of such a box if the amount of material used in its construction is to be minimal. Hint: Let the dimensions of the box be x by y by z.Then, xyz = 108 and the amount of material used is given by S = xy + 2yz + 2xz.Show that  Minimize f(x, y)
A)x = 1; y = 1; z = 2 
B)x = 2; y = 2; z = 1 
C)x = 2; y = 1; z = 1 
D)x = 2; y = 1; z = 2

Quizplus · Accepted Answer

Let's solve for z in terms of x and y using the given equation xyz = 108. 
z = 108/xy
Now substitute this value of z in the expression for the amount of material used S = xy + 2yz + 2xz
S = xy + 2xy(108/xy) + 2x(108/xy)
S = xy + 216/x + 216/y
To minimize S, we need to find the values of x and y that minimize this expression. One way to do this is to take partial derivatives with respect to x and y, set them equal to 0, and solve for x and y. Alternatively, we can use the fact that the minimum value of a sum is obtained when all terms are equal. This is known as the AM-GM inequality.
Using AM-GM, we have:
S/3 = (xy + 216/x + 216/y)/3 ≥ 3√(xy*216/x*216/y) = 36
S ≥ 108
The equality holds when xy = 216/x = 216/y, i.e., when x = y = 6. In this case, z = 108/xy = 3. Therefore, the dimensions of the box that minimize the amount of material used are x = y = 6 and z = 3. We can verify that this satisfies the volume constraint xyz = 108. Thus, the answer is B.

An Open Rectangular Box Having a Volume of Is