An open rectangular box having a volume of 108 in.3 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 in. by y in. by z in. Then, xyz

6 in x 6 in x 3 in.

Step-by-step explanation:

Given

V = xyz = 108   ⇒   z = 108/(xy)

The amount of the material used is

S = xy + 2yz + 2xz

Put value of z from the volume

S = xy + 2y*108/(xy) + 2x*108/(xy) = xy + 216/x + 216/y

Now, we find the relative minimum of the function S(x,y)

First, we find the critical point. Set Sx = 0  and  Sy = 0

and solve this system:

Sx(x,y) = y - (216/x²) = 0

Sy(x,y) = x - (216/y²) = 0

From the first equation we have

y = 216/x²

Put it in the second equation and find x

x - (216/(216/x²)²) = 0

⇒  x*(1 - (x³/216)) = 0

⇒  x₁ = 0   and  x₂ = 6

Now, we can find y as follows

y₁ = 216/(0)²   which is undefined

y₂ = 216/(6)² = 6

Hence, the only critical point of S is (6, 6). Next, we calculate the second ordered derivatives that we need for the second derivative test:

Sxx(x,y) = 432/x³

Sxy(x,y) = 1

Syy(x,y) = 432/y³

Applying the second derivative test

D(6, 6) = Sxx(6, 6)*Syy(6, 6) - S²xy(6, 6) = 2*2 - 1² = 4 -1 = 3 > 0

Sxx(6, 6) = 2 > 0

Since D(6, 6) > 0   and   Sxx(6, 6) > 0   we can conclude that S has a relative minimum at (6, 6).

z coordinate is:

z = 108/(xy) = 108 / (6*6) = 3

Finally, the dimentions of a box are 6 in x 6 in x 3 in.

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the  x-axis  . 

step-by-step explanation: