Finding the dimensions of a rectangular container?

**AlphaRock** · April 3rd 2010, 07:50 AM

What is the largest volume if the surface area of the container is 96 m^2?

My work:
V = lwh
V = xyz

SA = 96 = xy + 2xz 2yz
z = (96 - xy)/(2x + 2y)

V = xy(96 - xy)/(2x + 2y)
V = (96xy - (x^2)(y^2))/(2x + 2y)

Not sure if my work is on track... But can somebody show me how to get the answer? It's x = y = z = 4m.

**Chris L T521** · April 3rd 2010, 08:50 AM

Originally Posted by AlphaRock

What is the largest volume if the surface area of the container is 96 m^2?

My work:
V = lwh
V = xyz

SA = 96 = xy + 2xz 2yz
z = (96 - xy)/(2x + 2y)

V = xy(96 - xy)/(2x + 2y)
V = (96xy - (x^2)(y^2))/(2x + 2y)

Not sure if my work is on track... But can somebody show me how to get the answer? It's x = y = z = 4m.

From what you have in your SA equation, the container has no lid! If it has a lid, then $SA=96=2xy+2xz+2yz\implies z=\frac{48-xy}{x+y}$

Therefore, $V(x,y)=\frac{48xy-x^2y^2}{x+y}$ .

Now you have to find the critical point of this multivariable function.

Therefore, $\frac{\partial V}{\partial x}=\frac{(x+y)(48y-2xy^2)-(48xy-x^2y^2)}{(x+y)^2}=\frac{48y^2-x^2y^2-2xy^3}{(x+y)^2}$

and

$\frac{\partial V}{\partial y}=\frac{(x+y)(48x-2x^2y)-(48xy-x^2y^2)}{(x+y)^2}=\frac{48x^2-x^2y^2-2x^3y}{(x+y)^2}$

Now, set $\frac{\partial V}{\partial x}=0$ and $\frac{\partial v}{\partial y}=0$ . Observe that

$\frac{\partial V}{\partial x}=0\implies 48y^2-x^2y^2-2xy^3=0\implies x^2y^2=48y^2-2xy^3$

Therefore, $\frac{\partial V}{\partial y}=0\implies 48x^2-48y^2+2xy^3-2x^3y=0$

But this implies that $48(x^2-y^2)-2xy(x^2-y^2)=(48-2xy)(x^2-y^2)=0$ . Therefore $xy=24$ or $x=y$ .

If $x=y$ then $\frac{\partial V}{\partial x}=0\implies 48x^2-x^4-2x^4=0\implies$ $16x^2-x^4=0\implies x^2(16-x^2)=0$

Therefore, $x^2=0\implies x=0$ or $16-x^2=0\implies x=\pm 4$ . Since length can't be negative, we have $x=4$ .

As a result, $y=4$ . Therefore, $z=\frac{48-(4)(4)}{4+4}=\frac{32}{8}=4$ .

Therefore, the rectangular box that has maximum volume and SA = 96 sq. meters is a cube with dimensions $x=y=z=4$ meters.

You can verify that this is a maximum by applying the second derivative test at the point $(4,4,4)$ .

Does this make sense?

(I'm sure you could have applied Lagrange Multipliers here to $f(x,y,z)=xyz$ with the restriction $2xy+2xz+2yz=96$ (which could cut back on the calculations that are required), but this is what came to my mind first.)

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