A cylindrical package with half a hemisphere to be sent by a postal service can have a maximum combined length (height of the cylinder) and girth (perimeter of a cross section of the cylinder) of 108 inches. Find the dimensions of the package of maximum volume that can be sent. Assume the cross section is a circle. (Make sure you verify that the value you found is the maximum).

Radius of package is 36/π inches

Height of package is 36 inches

Step-by-step explanation:

Here we have

Sum of height plus perimeter of package = 108 inches that is

2πr + h = 108 inches

Where:

Height of package = h

Radius of package = r

Volume of package = πr²h

However, h = 108 - 2πr

Therefore, the equation for the volume of the package is;

Volume of package, V = πr² (108 - 2πr) = πr²*108 - 2π²r³

Differentiating the above equation and equating to zero to find a maximum value, we have;

2πr*108 - 6r²π² = 0

2πr*108 = 6r²π²

36 = r·π

r = 36/π inches

h = 108 - 2πr = 108 - 72 = 36 inches

Radius of package, r = 36/π inches

Height of package, h = 36 inches.

36 x 18 x 18

Step-by-step explanation:

By Lagrange multiplier's method,

∇f (x, y, z) = λ∇g (x, y, z) & g (x, y, z) = k

considering x, y, z as three unequal sides of the box

f (x, y, z) = xyz

If x represents the length, then constraint condition is g (x, y, z) = x + 2 (y+z) = 108

We have the following three equations:

yz = λ - ->eq (1) fx = λgx

xz=2λ - ->eq (2) fy=λgy

xy = 2λ - ->eq (3) fz=λgz

x+2y+2z=108-->eq (4)

Dividing eq (2) by eq (1), we'll have

x / y = 2

=> x = 2y

Also, by using eq (2) and eq (3), we can represent y=z

By substituting 'x = 2y' and 'y=z' in eq (4), we have

eq (4) = >

2y + 2 (y+y) = 108

y=108/6

y=18

For x: x=2y=> 2 (18) = >36

For z: z = y = 18

Therefore, the dimensions are 36 x 18 x 18