Answer to Question #249311 in Calculus for Anmol

Question

What are dimensions of rectangular prism with the largest possible volume that can be construed using 2400 square inches of surface area ?

Answer 1

If we are assuming a fully rectangular prism---right angles all around---then

$$2xy+2(x+y)z=2400$$

$$xy+(x+y)z=1200$$

$$z=\dfrac{1200-xy}{x+y}$$

$$V(x, y)=xy(\dfrac{1200-xy}{x+y})$$

$$V(x, y)=\dfrac{1200xy-x^2y^2}{x+y}$$

$$V_x=\dfrac{(1200y-2xy^2)(x+y)-(1200xy-x^2y^2)}{(x+y)^2}$$

$$=\dfrac{y^2(1200-x^2-2xy)}{(x+y)^2}$$

$$V_y=\dfrac{(1200x-2x^2y)(x+y)-(1200xy-x^2y^2)}{(x+y)^2}$$

$$=\dfrac{x^2(1200-y^2-2xy)}{(x+y)^2}$$

Find the critical (stationary) point(s)

$$\begin{matrix} V_x=0 \\ V_y=0 \end{matrix}$$

$$y^2(1200-x^2-2xy)=0$$

$$x^2(1200-y^2-2xy)=0$$

Then for $x>0, y>0$

$$x^2-y^2=0$$

$$(x-y)(x+y)=0$$

$$x=y$$

$$1200-x^2-2x^2=0$$

$$x^2=400, x>0$$

$$x=20=y$$

$$z=\dfrac{1200-20(20)}{20+20}=20$$

$$V=20(20)(20)=8000 (in^3)$$

$$V_{xx}=\dfrac{-2y^2(y^2+1200)}{(x+y)^3}$$

$$V_{yy}=\dfrac{-2x^2(x^2+1200)}{(x+y)^3}$$

$$V_{xy}=\dfrac{-2xy(x^2+y^2+3xy-1200)}{(x+y)^3}=V_{yx}$$

$$V_{xx}(20,20)=\dfrac{-2(20)^2((20)^2+1200)}{(20+20)^3}=-20<0$$

$$V_{yy}(20,20)=\dfrac{-2(20)^2((20)^2+1200)}{(20+20)^3}=-20$$

$$V_{xy}(20,20)=\dfrac{-2(20)(20)((20)^2+(20)^2+3(20)(20)-1200)}{(20+20)^3}$$

$$=-10$$

$$D=\begin{vmatrix} -20 & -10 \\ -10 & -20 \end{vmatrix}=300>0$$

The volume has a relative maximum at $(20, 20).$ Since there is the only stationary point for $x>0, y>0,$ then the volume has the absolute maximum for $x>0, y>0$ at $(20, 20).$

We have the cube with side $20$ inches.

$$V_{max}=20(20)(20)=8000\ in^3$$