# Danielle's question at Yahoo Answers regarding optimization

• MHB
• MarkFL
In summary: Registration is free and anonymous. In summary, at x=\frac{20\pm\sqrt{(-20)^2-4(3)(24)}}{2\cdot3}, the volume of the box is 68 cubic inches.
MarkFL
Gold Member
MHB
Here is the question:

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:

I have posted a link there to this topic so the OP can find my response.

Hello Danielle,

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

$$\displaystyle V=bh$$

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:

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

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

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

We should observe that we require $$\displaystyle 0<x<4$$ 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:

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

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

So, we find:

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

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

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

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

The root in the domain is:

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

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:

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

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:

## 1. What is optimization?

Optimization is the process of finding the best solution or value for a problem or situation. It involves maximizing or minimizing a certain parameter based on given constraints.

## 2. What is the context of Danielle's question about optimization on Yahoo Answers?

The context of Danielle's question is not specified, so it is difficult to provide a specific answer. However, optimization can be applied to various fields such as mathematics, engineering, computer science, economics, and more.

## 3. What are some common techniques used for optimization?

Some common techniques used for optimization include linear programming, gradient descent, genetic algorithms, and simulated annealing. These techniques use different approaches to find the optimal solution based on the given problem and constraints.

## 4. How is optimization different from maximization or minimization?

Optimization is a broader term that encompasses both maximization and minimization. Maximization refers to finding the highest possible value for a parameter, while minimization refers to finding the lowest possible value. Optimization, on the other hand, can involve both maximizing and minimizing, depending on the given problem.

## 5. Can optimization be applied to real-world problems?

Yes, optimization is a widely used technique in various industries to solve real-world problems. It can be applied to optimize production processes, financial investments, transportation routes, and more. It can also be used in decision-making to find the best course of action with limited resources.

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