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

#### MarkFL

Staff member
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:

CALCULUS needHELP WITH THIS PLEASE! APPRECIAT EIT -? - Yahoo! Answers

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

#### MarkFL

Staff member
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 Calculus forum.

Best Regards,

Mark.

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#### Prove It

##### Well-known member
MHB Math Helper
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*}.

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#### Prove It

##### Well-known member
MHB Math Helper
Sorry Mark, it appears you've forgotten that the OP stated the length is twice as long as the width