Question

1.Determine the nature of the stationary value of x= t^3 - 3t + ty^2

2. A rectangle box whose value is 32 is open at the top,if the surface area is 2(L+B)H + LB where LBH are Length,Breadth and Height Respectively

i.Find the Dimension of the box that may require list material

ii.investigate weather the dimension found requires List material.

Answer 1

1.

$$x(t, y)=t^3 - 3t + ty^2$$

$$\dfrac{\partial x}{\partial t}=3t^2-3+y^2$$

$$\dfrac{\partial x}{\partial y}=2yt$$

Find the critical point(s)

$$\dfrac{\partial x}{\partial t}=0,$$

$$\dfrac{\partial x}{\partial y}=0$$

$$3t^2-3+y^2=0$$

$$2yt=0$$

$$t=0, y=-\sqrt{3}$$

$$t=0, y=\sqrt{3}$$

$$y=0,t=-1$$

$$y=0, t=1$$

$$\dfrac{\partial^2 x}{\partial t^2}=6t$$

$$\dfrac{\partial^2 x}{\partial t\partial y}=2y$$

$$\dfrac{\partial^2 x}{\partial y^2}=2t$$

$$D=\begin{vmatrix} 6t & 2y \\ 2y & 2t \end{vmatrix}=12t^2-4y^2$$

$t=0, y=-\sqrt{3}$

$$\dfrac{\partial^2 x}{\partial t^2}|_{t=0, y=-\sqrt{3}}=0$$

$$D|_{t=0, y=-\sqrt{3}}=-12<0$$

$x(0, -\sqrt{3})$ is not a local maximum or local minimum.

$t=0, y=\sqrt{3}$

$$\dfrac{\partial^2 x}{\partial t^2}|_{t=0, y=\sqrt{3}}=0$$

$$D|_{t=0, y=\sqrt{3}}=-12<0$$

$x(0, \sqrt{3})$ is not a local maximum or local minimum.

$t=-1, y=0$

$$\dfrac{\partial^2 x}{\partial t^2}|_{t=-1, y=0}=-6<0$$

$$D|_{t=-1, y=0}=12>0$$

$x(-1, 0)$ is a local maximum.

$t=1, y=0$

$$\dfrac{\partial^2 x}{\partial t^2}|_{t=1, y=0}=6>0$$

$$D|_{t=1, y=0}=12>0$$

$x(1, 0)$ is a local minimum.

2.

Given

$$LBH=32=>H=\dfrac{32}{LB}$$

$$S=(2L+2B)H+LB$$

Substitute

$$S(L, B)=(2L+2B)(\dfrac{32}{LB})+LB$$

$$=\dfrac{64}{B}+\dfrac{64}{L}+LB, L>0, B>0$$

$$\dfrac{\partial S}{\partial L}=-\dfrac{64}{L^2}+B$$

$$\dfrac{\partial S}{\partial B}=-\dfrac{64}{B^2}+L$$

Find the critical point(s)

$$\dfrac{\partial S}{\partial L}=0,$$

$$\dfrac{\partial S}{\partial B}=0$$

$$-\dfrac{64}{L^2}+B=0$$

$$-\dfrac{64}{B^2}+L=0$$

$$BL^2=64=L^2B$$

$$B=L=4$$

$$\dfrac{\partial^2 S}{\partial L^2}=\dfrac{128}{L^3}$$

$$\dfrac{\partial^2 S}{\partial B^2}=\dfrac{128}{BL^3}$$

$$\dfrac{\partial^2 S}{\partial L\partial B}=1$$

$$D=\begin{vmatrix} 128/L^3 & 1 \\ 1 & 128/B^3 \end{vmatrix}=16384/(LB)^3-1$$

$$\dfrac{\partial^2 S}{\partial L^2}|_{L=4, B=4}=2>0$$

$$D|_{L=4, B=4}=4-1=3>0$$

$S(4, 4)$ is a local minimum.Since the function $S(L,B)$ has the only extreme value for $L>0, B>0,$ then S(4, 4) is the absolute minimm for $L>0, B>0.$

$$H=\dfrac{32}{4(4)}=2$$

The dimensions of the box for least material are

$$Length\times Breadth\times Height$$

$$=4\times4\times 2$$