# Thread: Surface area of cuboid

1. ## Surface area of cuboid

4 days till exams and counting down - nerves shattered already!
here goes:

A closed cuboid with rectangular base of width xcm. Length of base is twice the width and the volumn is 1944cm^3. The surface area of the cuboid is S cm^2.
a) Show that S = 4x^2 + 5832x^-1.
b) Given that x can vary, find the value of x that makes the surface area a minimum.
c) Find the minimum value of the surface area.

Going round in circles with this one. Or, going round in cuboids. ha ha
I kill me!

thanks all

2. Originally Posted by lemonz
4 days till exams and counting down - nerves shattered already!
here goes:

A closed cuboid with rectangular base of width xcm. Length of base is twice the width and the volumn is 1944cm^3. The surface area of the cuboid is S cm^2.
a) Show that S = 4x^2 + 5832x^-1.
b) Given that x can vary, find the value of x that makes the surface area a minimum.
c) Find the minimum value of the surface area.

Going round in circles with this one. Or, going round in cuboids. ha ha
I kill me!

thanks all
I presume that "cuboid", here, is what I would call a "rectangular solid".

Such a figure with "width" x, "length" y and "height" z has volume xyz and surface area 2xy+ 2xz+ 2yz.

For this particular figure you are told that "Length of base is twice the width" so y= 2x.

Also, "the volumn is 1944cm^3" so $xyz= x(2x)z= 2x^2z= 1944$ so that $z= 1944/2x^2$.

(a) Since S= surface area= 2xy+ 2xz+ 2yz, y= 2x and $z= 1994/2x^2$, $S= 2x(2x)+ 2x(1944/2x^2)+ 2(2x)(1944/2x^2)$ $= 4x^2+ 1944/x+ 3888/x= 4x^2+ 5832/x= 4x^2+ 5832/x$.

(b) One way to find the maximum value of a function is to find where its derivative is [/tex]0: S'= 8x- 5832/x^2= 0[/tex] so $8x^3= 5832$ and $x^3= 729$ and, finally, $x= ^3\sqrt{729}= 9 cm.$.

(c) $S(9)= 4(9)^2+ 5832/9= 324+ 648= 972 cm^2$

3. ## Thanx

Wow!
I have to say, that was a bit more than I expected. But, I worked my way through it and understand fully, now.
Thank you so much.
Regards