MHB Danielle's question at Yahoo Answers regarding optimization

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Optimization
AI Thread Summary
To determine the maximum volume of a gift box made from an 8 by 12 inch piece of cardboard, squares of side length x are cut from each corner. The volume function is defined as V(x) = 4x(6-x)(4-x), with valid x values between 0 and 4. By finding the derivative and applying the quadratic formula, the critical point is x = (2(5 - √7))/3. Evaluating the volume at this critical value yields approximately 68 cubic inches, which is the maximum volume achievable. This solution encourages further exploration of optimization problems in related forums.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Helpppppppppppp please !?

you are making the bottom of a gift box from a piece of carboard that is 8 inches by 12 inches. to the nearest of a cubic inch, what is the maximum volume of the box?

Here is a link to the question:

Helpppppppppppp please !? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello Danielle,

The volume of the box is the product of the area of its base and its height:

$$V=bh$$

We will be cutting squares, let's let the side length of these squares be $x$, from each corner of the cardboard sheet, so that we have flaps to fold up to make the box bottom. Hence, the area of the base is:

$$b=(12-2x)(8-2x)=4(6-x)(4-x)$$

and the height of the box is $x$, and so we have:

$$V(x)=4x(6-x)(4-x)$$

We should observe that we require $$0<x<4$$ in order to get meaningful results.

Next, we need to equate the derivative of the volume function to zero, and look at critical values in the valid domain. We will use the following rules of differentiation:

$$g(x)=k\cdot f(x)\implies g'(x)=k\cdot f'(x)$$ where $k$ is a constant.

$$u(x)=f(x)g(x)h(x)\implies u'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$$

So, we find:

$$V'(x)=4\left((1)(6-x)(4-x)+x(-1)(4-x)+x(6-x)(-1) \right)=$$

$$4\left(24-10x+x^2-4x+x^2-6x+x^2 \right)=4(3x^2-20x+24)=0$$

Applying the quadratic formula, and taking the root in the domain, we find:

$$x=\frac{20\pm\sqrt{(-20)^2-4(3)(24)}}{2\cdot3}=\frac{20\pm4\sqrt{7}}{6}=\frac{2(5\pm\sqrt{7})}{3}$$

The root in the domain is:

$$x=\frac{2(5-\sqrt{7})}{3}$$

To verify this critical value is at a maximum, we may use the first derivative test, and observe that it is positive to the left of this critical value and negative to the right, hence the critical value is at a maximum.

Now, to find the volume of the box at this value of $x$, we need to evaluate:

$$V\left(\frac{2(5-\sqrt{7})}{3} \right)=\frac{64(7\sqrt{7}+10)}{27}\approx68\text{ in}^3$$

To Danielle and any other visitors viewing this topic, I encourage you to register and post other optimization problems in our http://www.mathhelpboards.com/f10/ forum.
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top