Zhina's question at Yahoo Answers regarding optimization subject to constraint

• MHB
• MarkFL
In summary: Thank you for correcting me.In summary, Mark has provided a summary of the content. The question asked for help with a calculus problem, and the summary states that the maximum volume for a box can be found when the width is twice the length and the height is six-thirds of the width.
MarkFL
Gold Member
MHB
Here is the question:

CALCULUS needHELP WITH THIS PLEASE! APPRECIAT EIT -?

please DON'T GO AWAY WHEN YOU SEE THIS! HELP NEEDED THANKS

a 24 foot wire is cut into 12 pieces which are welded together to form a rectangular frame whose base is twice as long than it is wide .
the frame is than covered with paper to form a box . find the volume of the largest possible box that can be made this way

Here is a link to the question:

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

Hello Zhina,

Let's let the dimensions of the box be $x,y,z$ (all positive) where the volume (the objective function) is:

$$\displaystyle V(x,y,z)=xyz$$

Each dimension will require 4 pieces of wire, so we have the constraint:

$$\displaystyle 4x+4y+4z=4L$$ (where $4L$ is the total length of the wire)

$$\displaystyle g(x,y,z)=x+y+z-L=0$$

We also have the constraint (if we choose $y$ to be the length, and $x$ to be the width)

$$\displaystyle h(x,y,z)=y-2x=0$$

Using Lagrange multipliers, we obtain the system:

$$\displaystyle yz=\lambda-2\mu$$

$$\displaystyle xz=\lambda+\mu$$

$$\displaystyle xy=\lambda$$

which implies:

$$\displaystyle \lambda=yz+2\mu=xz-\mu=xy$$

From:

$$\displaystyle yz+2\mu=xz-\mu$$

we obtain:

$$\displaystyle \mu=\frac{xz-yz}{3}$$

Now we also have:

$$\displaystyle xz-\mu=xy$$

$$\displaystyle \mu=xz-xy$$

and hence, we find:

$$\displaystyle \frac{xz-yz}{3}=xz-xy$$

$$\displaystyle xz-yz=3xz-3xy$$

$$\displaystyle -yz=2xz-3xy$$

$$\displaystyle x=\frac{yz}{3y-2z}$$

Substituting for $x$ into the second constraint, we find:

$$\displaystyle y=\frac{2yz}{3y-2z}$$

Since $y\ne0$, we may divide through by $y$ to obtain:

$$\displaystyle y=\frac{4z}{3}$$

Substituting for $y$ into the formula for $x$, we obtain:

$$\displaystyle x=\frac{\left(\frac{4z}{3} \right)z}{3\left(\frac{4z}{3} \right)-2z}=\frac{2z}{3}$$

Now, substituting for $x$ and $y$ into the first constraint, we obtain:

$$\displaystyle \frac{2z}{3}+\frac{4z}{3}+z=L$$

$$\displaystyle z=\frac{L}{3},\,y=\frac{4L}{9},\,x=\frac{2L}{9}$$

We know this is a maximum, as the other critical point is $$\displaystyle \left(0,0,4L \right)$$ giving:

$$\displaystyle V_{\min}=0\text{ ft}^3$$

Thus, we find:

$$\displaystyle V_{\max}=V\left(\frac{2L}{9},\frac{4L}{9},\frac{L}{3} \right)=3\left(\frac{2L}{9} \right)^3$$

Plugging in the given $$\displaystyle L=6\text{ ft}$$, we find:

$$\displaystyle V_{\max}=3\left(\frac{2(6\text{ ft})}{9} \right)^3=\frac{64}{9}\,\text{ ft}^3$$

To Zhina and any other guests viewing this topic, I invite and encourage you to post other optimization questions in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

Last edited:
This can actually be solved using single variable calculus. We know that the twelve pieces form the edges of a box, and we also know that we are making these pieces from a 24ft long wire. If we call the length "l", the width "w" and the height "h", then we have

\displaystyle \displaystyle \begin{align*} 4l + 4w + 4h &= 24 \end{align*}

We are also told that the length is twice the width, so \displaystyle \displaystyle \begin{align*} l = 2w \end{align*}, and from there we can see that

\displaystyle \displaystyle \begin{align*} 4 \left( 2w \right) + 4w + 4h &= 24 \\ 12w + 4h &= 24 \\ 4h &= 24 - 12w \\ h &= 6 - 3w \end{align*}

The volume of the box can be found using

\displaystyle \displaystyle \begin{align*} V &= lwh \\ &= 2w \left( w \right) \left( 6 - 3w \right) \\ &= 12w^2 - 6w^3 \end{align*}

The maximum volume is where the derivative is 0, so

\displaystyle \displaystyle \begin{align*} \frac{dV}{dw} &= 24w - 18w^2 \\ 0 &= 24w - 18w^2 \\ 0 &= 6w \left( 4 - 3w \right) \\ w = 0 \textrm{ or } w &= \frac{4}{3} \end{align*}

Clearly when w = 0, the volume will be 0, so the maximum volume has to be when \displaystyle \displaystyle \begin{align*} w = \frac{4}{3} \end{align*}. To check that it's a maximum, we can check the value of the derivative on either side of the point (say at w = 1 and w = 2). So

\displaystyle \displaystyle \begin{align*} \frac{dV}{dw} | _{w = 1} &= 6 \\ \\ \frac{dV}{dw} | _{w = 2} &= -24 \end{align*}

Since the gradients go from positive to 0 to negative, the criticial point is a maximum.

\displaystyle \displaystyle \begin{align*} V &= 12 \left( \frac{4}{3} \right) ^2 - 6 \left( \frac{4}{3} \right) ^3 \\ &= 12 \left( \frac{16}{9} \right) - 6 \left( \frac{64}{27} \right) \\ &= \frac{64}{3} - \frac{128}{9} \\ &= \frac{64}{9} \end{align*}

Therefore the maximum volume is \displaystyle \displaystyle \begin{align*} \frac{64}{9} \,\textrm{ft}\,^3 \end{align*}, when the box has dimensions \displaystyle \displaystyle \begin{align*} \frac{8}{3} \,\textrm{ft} \, \times \frac{4}{3} \,\textrm{ft} \, \times 2\,\textrm{ft} \end{align*}.

Last edited:
Sorry Mark, it appears you've forgotten that the OP stated the length is twice as long as the width :)

Prove It said:
Sorry Mark, it appears you've forgotten that the OP stated the length is twice as long as the width :)

Yes, thanks to your post, I realized I did not read the problem carefully enough, and I have edited my post to reflect the second constraint. I would not have used Lagrange multipliers if I had done so. :D

Last edited:

1. What is optimization subject to constraint?

Optimization subject to constraint is a mathematical approach to finding the best solution for a problem while taking into account certain limitations or restrictions. It involves maximizing or minimizing a function while adhering to certain constraints.

2. Why is optimization subject to constraint important?

Optimization subject to constraint is important because it allows us to find the most efficient or effective solution to a problem while considering real-world limitations. This can be applied in various fields such as engineering, economics, and operations research.

3. How is optimization subject to constraint different from regular optimization?

Regular optimization involves finding the best solution for a problem without any restrictions or limitations. On the other hand, optimization subject to constraint takes into account real-world constraints, making it more applicable to practical situations.

4. What are some common techniques used in optimization subject to constraint?

Some commonly used techniques in optimization subject to constraint include linear programming, quadratic programming, and nonlinear programming. Other methods such as genetic algorithms and simulated annealing can also be used to solve more complex problems.

5. What are some real-life applications of optimization subject to constraint?

Optimization subject to constraint has numerous applications in various fields. For example, in manufacturing, it can be used to optimize production processes while adhering to resource and time constraints. In finance, it can be used to optimize investment portfolios while considering risk and return. It is also commonly used in transportation, logistics, and supply chain management to optimize routes and schedules while considering factors such as cost and time.

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