A box with a square base and open top must have a volume of 32,000cm^3. How do you find the dimensions of the box that minimize the amount of material used?

Ariel Wilkinson 2022-09-06 Answered
A box with a square base and open top must have a volume of 32,000cm^3. How do you find the dimensions of the box that minimize the amount of material used?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

beshrewd6g
Answered 2022-09-07 Author has 12 answers
The Volume of a box with a square base x by x cm and height h cm is V = x 2 h
The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.
The surface area of the box described is A = x 2 + 4 x h
We need A as a function of x alone, so we'll use the fact that V = x 2 h = 32 , 000 cm^3
which gives us h = 32 , 000 x 2 , so the area becomes:
A = x 2 + 4 x ( 32 , 000 x 2 ) = x 2 + 128 , 000 x
We want to minimize A, so
A = 2 x - 128 , 000 x 2 = 0 when 2 x 3 - 128 , 000 x 2 = 0
Which occurs when x 3 - 64 , 000 = 0 or x=40
The only critical number is x=40 cm.
The second derivative test verifies that A has a minimum at this critical number: A = 2 + 256 , 000 x 3 which is positive at x=40.
The box should have base 40 cm by 40 cm and height 20 cm.
(use h = 32 , 000 x 2 and x=40)
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-05-09
Most statements of Constraint Qualification I have found in the literature mention a locally "locally optimal solution" of the problem:
{ min f ( x ) s.t. g i ( x ) 0
It is stated that when a C.Q. holds at a local optimum, then there exist Lagrange multipliers that satisfy KKT conditions.
But, I cannot get my head around this notion of local optimality. Does it mean locally optimal for the unconstrained problem? Does not local optimality imply the satisfaction of the KKT conditions?
asked 2022-10-18
A rectangular box is to be inscribed inside the ellipsoid 2 x 2 + y 2 + 4 z 2 = 12 . How do you find the largest possible volume for the box?
asked 2022-10-06
How do you find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum?
asked 2022-08-13
How do you find the point on the curve y = 2 x 2 closest to (2,1)?
asked 2022-09-23
How do you find the length and width of a rectangle whose area is 900 square meters and whose perimeter is a minimum?
asked 2022-08-11
How do you find the dimensions that minimize the amount of cardboard used if a cardboard box without a lid is to have a volume of 8 , 788 ( c m ) 3 ?
asked 2022-06-20
Find global extrema of f ( x , y ) = x 2 2 + 3 y 2 on the set M ( g ) := { ( x , y ) R 2 g ( x , y ) = 0 } where g ( x , y ) = x 2 + y 4 25.

M ( g ) is compact and f continuous so we know that there must exists a global maximum and global minimum. The conditions of the Lagrange multiplier methods are satisfied and if we solve the equations that result from the Lagrange multiplier method we get the following points:
( 0 , 5 ) , ( 0 , 5 ) , ( 5 , 0 ) , ( 5 , 0 ) , ( 4 , 3 ) , ( 4 , 3 ) , ( 4 , 3 ) , ( 4 , 3 ) .
and
f ( 0 , 5 ) = 15 , f ( 0 , 5 ) = 15 , f ( 5 , 0 ) = 25 2 , f ( 5 , 0 ) = 25 2 , f ( 4 , 3 ) = 17 , f ( 4 , 3 ) = 17 , f ( 4 , 3 ) = 17 , f ( 4 , 3 ) = 17.
So far so good.

However our sample solution says that "from the above values we see that 25 2 is the global minimum and 17 the global maximum. "

As far as I have understood the Lagrange multiplier method it only delivers a necessary but not sufficient condition. So we don't know if one of the three points 25 2 , 15 , 17 is a saddle point. To make sure that the points are indeed extrema we have to resort to another method (e.g. plug in the condition into f).

Am I am right or is there something I don't see or didn't understand correctly?