Box Optimization problem

In summary: You are solving 12h^2- 112h+ 180= 0 by factoring 4 out of the first two terms: 4(3h^2- 28h+ 45)= 0. Now you can see that 3h^2- 28h+ 45= 0 is a quadratic equation in h and can be factored as (3h- 5)(h- 9)= 0. That will give you three values for h: 3h- 5= 0 gives h= 5/3 while h- 9= 0 gives h= 9. Of course, h= 5/3 will give a
  • #1
computerex
68
0

Homework Statement



Question:

You are planning to make an open rectangular box from a 10 by 18 cm piece of cardboard by cutting congruent squares from the corners and folding up the sides.

1) What are the dimensions of the box of largest volume you can make this way?
2) What is its volume?

v - volume
l - length
h - height
w - width

Homework Equations



v = lwh

The Attempt at a Solution



Since v = lwh w = 10 - 2h and l = 18 - 2h.
Therefore:

v(h) = (18-2h)(10-2h)(h)
v(h) = (4h^3) - (56h^2) + 4h^2

Differentiation:

dv/dh = (12h^2) - 112h + 180

Finding critical numbers:

(12h^2) - 112h + 180 = 0
(12h^2) - 112h = -180
h(12h - 112) = -180

12h - 112 = -180
12h = -68
h = -68/12

critical numbers: k{-180, -68/12}

My question is, how can I have a negative height? Is this correct so far or have I made some error?
 
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  • #2
for the purpose of this problem you cannot have negative heights, you made a simple error. You can fix it easily. the problem is your step near the end when you have
(12h^2) - 112h + 180 = 0 don't subtract the 180. when you get to this part factor our a value of 4 you should get 4((3h^2)-28h+45) from here just use the quadratic formula to find your zero's. you don't need to factor out the 4 you can just go straight to the quadratic formula, but it makes it less tedious if you do. =) Study Hard.
 
  • #3
ab= 180 does NOT mean "a= 180" or "b= 180" as you seem to think. If ab= 0, then either a= 0 or b= 0 but that is a specific property of 0 not of other numbers.
 

What is the "Box Optimization Problem"?

The Box Optimization Problem is a mathematical problem that involves finding the optimal dimensions of a box in order to maximize or minimize a certain objective function. It is often used in fields such as operations research, engineering, and computer science.

What are the variables involved in the Box Optimization Problem?

The variables in the Box Optimization Problem typically include the dimensions of the box (length, width, and height) as well as any constraints or limitations for these dimensions, such as a maximum volume or weight.

What are some real-life applications of the Box Optimization Problem?

The Box Optimization Problem can be applied to various real-life situations, such as determining the optimal dimensions for shipping boxes to minimize shipping costs, finding the most efficient layout for packaging products in a warehouse, and designing the dimensions of a container to maximize its capacity.

What methods are commonly used to solve the Box Optimization Problem?

There are various methods that can be used to solve the Box Optimization Problem, including linear programming, dynamic programming, and gradient descent algorithms. The choice of method depends on the specific problem and its constraints.

What are the limitations of the Box Optimization Problem?

The Box Optimization Problem relies on certain assumptions and simplifications, such as assuming a constant density and uniform shape of the objects being packed. In real-life scenarios, these assumptions may not hold true, which can affect the accuracy of the optimization results.

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