Question

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).

Answer 1

Radius of package is 36/π inches

Height of package is 36 inches

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.

Answer 2

36 x 18 x 18

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