# Min/Max Surface Area Cubes (chk anwser)

• Aug 8th 2006, 01:59 PM
Yogi_Bear_79
Min/Max Surface Area Cubes (chk anwser)
A mass of clay of volume $\displaystyle 432 in.^3$is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

(A) The minimum total surface area is obtained by making one cube as small as possible, $\displaystyle .01^3$, and the second cube $\displaystyle 7.5595^3$. The combined total surface area of these two cubes is (0.06+342.8) = 342.86.

The maximum total surface area can be reached with two equal cubes. $\displaystyle 6^3$, for each cube. The combined total surface area of these two cubes is $\displaystyle (216^2)$ = 432.
• Aug 8th 2006, 02:16 PM
Quick
Quote:

Originally Posted by Yogi_Bear_79
A mass of clay of volume $\displaystyle 432 in.^3$is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

(A) The minimum total surface area is obtained by making one cube as small as possible, $\displaystyle .01^3$, and the second cube $\displaystyle 7.5595^3$. The combined total surface area of these two cubes is (0.06+342.8) = 342.86.

The maximum total surface area can be reached with two equal cubes. $\displaystyle 6^3$, for each cube. The combined total surface area of these two cubes is $\displaystyle (216^2)$ = 432.

A much different approach:

volume of a cube is: $\displaystyle s^3$

surface area of a cube is: $\displaystyle 6s^2$

so then we get $\displaystyle V_1+V_2=432$

so then we say that: $\displaystyle V_2=432-V_1$

then rewrite: $\displaystyle s_2^3=432-s_1^3$

therefore: $\displaystyle s_2=\sqrt[3]{432-s_1^3}$

we're trying to find surface area (SA) so we say: $\displaystyle SA=SA_1+SA_2$

therefore: $\displaystyle SA=6s_1^2+6\sqrt[3]{432-s_1^3}$

you can then evaluate the answers from that.
• Aug 8th 2006, 03:59 PM
Soroban
Hello, Yogi_Bear_79!

Since this is posted under Calculus, I assume we're supposed to do Calculus.

Quote:

A mass of clay of volume 432 in³ is formed into two cubes.
What is the minimum possible total surface area of the two cubes?
What is the maximum?

Let $\displaystyle x$ and $\displaystyle y$ be the sides of the two cubes.

Their volumes satisfy: .$\displaystyle x^3 + y^3\:=\:432\quad\Rightarrow\quad y^3\:=\:432-$$\displaystyle x^3\quad\Rightarrow\quad y^2\:=\:\left(432-x^3\right)^{\frac{2}{3}}$ [1]

Their total surface area is: .$\displaystyle A\;=\;6x^2 + 6y^2$ [2]

Substitute [1] into [2]: .$\displaystyle A \;= \;x^3 + 6\left(432 - x^3\right)^{\frac{2}{3}}$

Differentiate: .$\displaystyle A' \;= \;12x + 6\cdot\frac{2}{3}\left(432-x^3\right)^{-\frac{1}{3}}(-3x^2)$

And we have: .$\displaystyle 12a - \frac{12x^2}{(432-x^3)^{\frac{1}{3}}} \;=\;0$

Multiply by $\displaystyle (432 - x^3)^{\frac{1}{3}}:\;\;12x(432 - x^3)^{\frac{1}{3}} - 12x^2\;=\;0$

Factor: .$\displaystyle 12x\left[(432 - x^3)^{\frac{1}{3}} - x\right] \;= \;0$

We have two equations to solve:

. . $\displaystyle 12x \,= \,0\quad\Rightarrow\quad \boxed{x \,= \,0}$

. . $\displaystyle (432-x^3)^{\frac{1}{3}} - x\:=\:0\quad\Rightarrow\quad(432-x^3)^{\frac{1}{3}}\:=$ $\displaystyle \:x\quad\Rightarrow\quad 432 - x^3\:=\:x^3$

. . . . . . . . $\displaystyle \Rightarrow\quad 2x^3\:=\:432\quad\Rightarrow\quad x^3\:=$ $\displaystyle \:216\quad\Rightarrow\quad \boxed{x\,=\,6}$

$\displaystyle x = 0$ means that the clay is formed into one cube.
. . Its surface area is: .$\displaystyle A\:=\:6(432)^{\frac{2}{3}} \:\approx\:342.88$ in² . . . minimum

$\displaystyle x = 6$ means the clay is formed into two equal cubes.
. . Their surface area is: .$\displaystyle A\:=\:6^3 + 6(216)^{\frac{2}{3}}\:=\:432$ in² . . . maximum

• Aug 8th 2006, 04:23 PM
Quick
Here's the graph of the equation:
(it looks as though there are 2 minimum values)
• Aug 8th 2006, 04:39 PM
Yogi_Bear_79
Thank you both for your help. I have a question. The original question states that the clay is formed into two cubes. So wouldn't that make Sorban's minimum anwser incorrect for the question?
• Aug 8th 2006, 06:24 PM
Soroban
Hello again, Yogi_Bear_79!

Quote:

The original question states that the clay is formed into two cubes.
So wouldn't that make Sorban's minimum answer incorrect for the question?

I thought about that . . . and had to make a decision.

If one cube is infinitesimally small, the total surface area approaches $\displaystyle 6(432)^{\frac{2}{3}}$
. . Hence, there is no minimum surface area.

So I assumed that one cube of side 0 is allowed.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

There are two minimums because there are two solutions:
. . $\displaystyle (x,y) \,=\,(0,6)$ and $\displaystyle (6,0).$

• Aug 8th 2006, 06:35 PM
ThePerfectHacker
It makes more sense to say that there is not minimum because the interval is open. And thus you do not include the endpoints.
• Aug 8th 2006, 07:01 PM
JakeD
Quote:

Originally Posted by Yogi_Bear_79
A mass of clay of volume $\displaystyle 432 in.^3$is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

(A) The minimum total surface area is obtained by making one cube as small as possible, $\displaystyle .01^3$, and the second cube $\displaystyle 7.5595^3$. The combined total surface area of these two cubes is (0.06+342.8) = 342.86.

The maximum total surface area can be reached with two equal cubes. $\displaystyle 6^3$, for each cube. The combined total surface area of these two cubes is $\displaystyle (216^2)$ = 432.

There is a minimum allowed length of .01 of the side of a cube. So two cubes are required, the allowed lengths form a closed interval and a minimum exists. But that requires some change to Soroban's proof to show that the minimum surface area occurs at an endpoint. Quick's graph shows this but some proof is probably necessary for the exercise.