A box, open at the top is to be made from cardboard. The base of the box is a square of side x and its height is y. If the volume of the box is 32u^2, find the dimensions of the box if the area is to be least. Please help?

2 Answers
May 29, 2018

See the explanation please.

Explanation:

The indicated volume does not make any sense and needs to be
clarified, but for now we can assume the volume to be #32 m^3#:

The area of the base is:

# A_(base)=x^2#

The volume would be:

#V=x^2y=32m^3#

Calculating #y# in terms of #x# in above equation results:

#y=32/x^2#

Total surface area of a box with open top is:

#A=x^2+4xy#

Substituting for #y#:

#A=x^2+4x * 32/x^2#

#A=x^2+128/x#

To find the critical points equate the first derivative to zero:

#(dA)/dx=A'=2x-128/x^2#

#(2x^3 - 128)/x^2=0#

#x^3-64=0#

#x^3=64#

#x=4m#

To verify the nature of the critical point use 2nd derivative test:

#A"=2+256/x^3#

#A"(4)=2+256/4^3=6>0=># Verifies it is a minimum so:

#x=4m#

#y=32/16=2m#

Thus the box dimensions for a minimum surface area are:

#4m * 4m * 2m#

And the minimum surface area would be:

#A=x^2+4xy=4^2+4*4*2=16+32=48m^2#

May 30, 2018

#x=4root3[u^2#............ #y=2root3[u^2#

Explanation:

Assuming the units expressed for the volume are correct and are indeed expressed in terms of #u^2 # then we may say ,

Volume #V=x^2y=32u^2#........#[1]# , the surface area #A=x^2+4xy#......#[2]#.

From ....#[1]# #y=32u^2/x^2# and substituting this value for #y# in .....#[2]#,

#A=x^2+4x[32u^2]/[x^2]# = #[x^2+[4][32u^2]]/x]#......#[3]#

Differentiating ......#[3]# wrt #x#, #d/dx#= #2x - 4[32u^2]/x^2# =0 [ for max/min and considering #u # constant]

#x^3 = 64u^2#, therefore #x = 4root3[u^2#. substituting this value for #x# in .....#[1]# will give the value of #y# in the answer.

The second derivative =#2+8[32u^2]/x^3# which is positive, since #x# and #u# must be both positive [ must have positive values for a box] and so confirms that the value of #x# obtained will minimise the surface area of the box.