# Find the minimize volume possible for the box v = xyz with an open top and the surface area of 12...

## Question:

Find the minimize volume possible for the box {eq}v = xyz {/eq} with an open top and the surface area of 12 square meter. partial derivatives

## Lagrangian Multiplier:

Given a function {eq}\displaystyle f(x,y,z) {/eq} subject to the constraint, {eq}\displaystyle g(x,y,z)=c {/eq} we assume a multiplier {eq}\displaystyle \lambda {/eq} to get a function {eq}\displaystyle F(x,y,z,\lambda)=f(x,y,z)-\lambda(g(x,y,z)-c) {/eq} and then we calculate the partial differentiation w.r.t. to each variable and equate it to zero as,

{eq}\displaystyle F_x=0,F_y=0,F_z\text{ and }F_{\lambda}=0 {/eq}

Solving the four equations we get the value of {eq}\displaystyle \lambda {/eq} and then we find the values of {eq}\displaystyle x,y\text{ and }z {/eq}.

The volume of the box is given to be, {eq}\displaystyle V=xyz {/eq} so the dimensions of the box will be {eq}\displaystyle x,y,\text{ and }z {/eq}. Let us assume that z is the height of the box. Considering this, the surface area of the open box will be,

{eq}\displaystyle \begin{align} A&=2(x\times z)+2(y\times z)+(x\times y)\\ &=xy+2xz+2yz \end{align} {/eq}

But the surface area is given to be 12. So the given problem reduces to,

We will use Lagrangian multiplier to solve this problem,

{eq}\displaystyle \text{ Let } F(x,y,z.\lambda)=xyz-\lambda(xy+2xz+2yz-12) {/eq}

So we will find the partial derivatives of w.r.t. to all the variables and set each of them to 0

{eq}\displaystyle \begin{align} F_x(x,y,z,\lambda)&=yz-\lambda(y+2z)=0\Rightarrow yz=\lambda(y+2z) \therefore \ xyz=\lambda x(y+2z)......(1)\\ \\ F_y(x,y,z,\lambda)&=xz-\lambda(x+2z)=0\Rightarrow xz=\lambda(x+2z) \therefore \ xyz=\lambda y(x+2z)......(2)\\ \\ F_z(x,y,z,\lambda)&=xy-\lambda(2x+2y)=0\Rightarrow xy=\lambda(2x+2y) \therefore \ xyz=\lambda z(2x+2y)......(3)\\ \\ F_{\lambda}(x,y,z,\lambda)&=(xy+2xz+2yz-12)=0\Rightarrow xy+2xz+2yz=12......(4)\\ \end{align} {/eq}

Solving the equations (1) and (2) we have,

{eq}\displaystyle \begin{align} &\lambda x(y+2z)=\lambda y(x+2z)\\ &\lambda x(y+2z)-\lambda y(x+2z)=0\\ &\lambda(2xz-2yz)=0\\ &2\lambda (xz-yz)=0\\ &\lambda=0 \text{ or }xz=yz \end{align} {/eq}

Here {eq}\displaystyle \lambda = 0 {/eq} simply means that from the equation set we get {eq}\displaystyle x=0\text{ or }y=0\text{ or }z=0 {/eq} which cant be possible as neither of the dimensions can be zero, else it will give us a zero volume.

So we get, {eq}\displaystyle x=y {/eq}

Also solving the equations (2) and (3) we get,

{eq}\displaystyle \begin{align} &\lambda y(x+2z)=\lambda z(2x+2y)\\ &\lambda y(x+2z)-\lambda z(2x+2y)=0\\ &\lambda(xy-2xz)=0\\ &2\lambda (xy-2xz)=0\\ &\lambda=0 \text{ or }xy=2xz \end{align} {/eq}

Again {eq}\displaystyle \lambda = 0 {/eq} is not valid. So we get, {eq}\displaystyle y=2z {/eq}. Thus finally we have the solution as, {eq}\displaystyle x=y=2z {/eq}

Now putting these values in the equation (4) we get,

{eq}\displaystyle \begin{align} &xy+2xz+2yz=12\\ \Rightarrow &(2z)(2z)+2(2z)z+2(2z)(z)=12\\ \Rightarrow &4z^2+4z^2+4z^2=12\\ \Rightarrow &12z^2=12\\ \Rightarrow &z^2=1\\ \Rightarrow &z=1 \text{ and }x=y=2(1)=2 \end{align} {/eq}

Finally lets find the volume , it will be,

{eq}\displaystyle \begin{align} V=(2)(2)(1)=4 \end{align} {/eq}

Now if we consider some other value of {eq}\displaystyle x=y=1 {/eq} then the z will be

{eq}\displaystyle (1)(1)+2(1)z+2(1)z=12\Rightarrow z=\frac{11}{4} {/eq}.

For these values the volume will be,

{eq}\displaystyle V=(1)(1)\left( \frac{11}{4}\right)=\frac{11}{4}\lt 4 {/eq}

As the new values of x and y give a volume which is less than the value 4 , hence the solution {eq}\displaystyle x=y=2 \text{ and }z=1 {/eq} gives us a maxima.

[Note: the minima will be x=y=z=0, which will give a zero volume, but this makes no sense as there is no box to consider for that solution.]