Surface are a of two cubes

[confused_07]

A mass of clay of V=432 cu.in. is formed into two cubes. What is the minimum possible total surface area of the two cubes? What is the maximum?

So far I know:

V=x^2y
Surface= 6(x^2)

How do I even start this? Do I set d(s)/d(x) for surface?

 

[soroban]

Hello, confused_07!

This is not a simple problem . . .

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

Let \(\displaystyle x\) anmd \(\displaystyle y\) be the sides of the two cubes.
Their volumes are \(\displaystyle x^3\) and \(\displaystyle y^3\).
Their surface areas are \(\displaystyle 6x^2\) and \(\displaystyle 6y^2\).

We are told that: \(\displaystyle \:x^3\,+\,y^3\:=\:432\;\) [1]

We want to maximize/minimize: \(\displaystyle \:A\:=\:6x^2\,+\,6y^2\;\) [2]

From [1], we have: \(\displaystyle \:y^3\:=\:432\,-\,x^3\)

Raise both sides to the \(\displaystyle \frac{2}{3}\) power: \(\displaystyle \:y^2\:=\:\left(432\,-\,x^3)^{\frac{2}{3}}\)

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

Differentiate and equate to zero:

. . \(\displaystyle \L A' \;=\;12x\,+\,6\,\cdot\,\frac{2}{3}(432\,-\,x^3)^{-\frac{1}{3}}(-3x^2) \:=\:0\)

. . and we have: \(\displaystyle \L\:12x \,-\,\frac{12x^2}{(432\,-\,x^3)^{\frac{1}{3}}} \;=\;0\)

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

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


And we have two equation to solve:

. . \(\displaystyle 12x\:=\:0\;\;\Rightarrow\;\;\fbox{x\:=\:0}\;\) [3]

. . \(\displaystyle (432\,-\,x^3)^{\frac{1}{3}}\,-\,x\:=\:0\;\;\Rightarrow\;\;(432\,-\,x^3)^{\frac{1}{3}}\:=\:x\)
. . Cube both sides: \(\displaystyle \:432\,-\,x^3\:=\:x^3\;\;\Rightarrow\;\;2x^3\:=\:432\;\;\Rightarrow\;\;\fbox{x\:=\:6}\;\) [4]

If [3] \(\displaystyle x\,=\,0\), substitute into [1] and get: \(\displaystyle \,y\,=\,6\sqrt[3]{2}\)

If [4] \(\displaystyle x\,=\,6\), substitute into [1] and get: \(\displaystyle \,y\,=\,6\)


We have two critical points: \(\displaystyle \(0,\,6\sqrt[3]{2}),\;Q(6,\,6)\)

With \(\displaystyle P(0,\,6\sqrt[3]{2})\), we use the clay to form one cube.
Substitute into [2]: \(\displaystyle \:A\:=\:6\cdot0^2\,+\,6(6\sqrt[3]{2})^3\:\approx\:\L342.9\) in² . . . minimum area

With \(\displaystyle Q(6,\,6)\), we use the clay to form two equal cubes.
Substitute into [2]: \(\displaystyle \:A\:=\:6\cdot6^2\,+\,6\cdot6^2\:=\:\L432\) in² . . . maximum area

 

[confused_07]

Wow!!!! I am really going to have to get the hang of how you figure things out like that..... Thank you.

 

[confused_07]

Is there a mistake at the end? Shouldn't it be

A= 6*(0)^2 + 6(6cube root(2))^2

Just making sure....