Maximizing Volume of a Cuboid - Nthabiseng P's Q&A

[MarkFL]

Here is the question:

Mathematics -derivatives?

Hi

Can someone help me to solve the following problem please

A box in the shape of a cuboid has dimensions:
Height: 50 cm
Width: 30 cm
Height: 20 cm
a) Calculate the total transmission restriction area and volume. ( i have done this)

need help with number b

b) Construct a new box with the same restriction area as the first box, but with different dimensions. Which dimensions of the box (length, width and height) give the largest possible volume?

(Hint: second box smallest side will be square. Calling the short side length of x and the long side length of y)Here's how far I've come:

a)
V = 20 * 30 * 50 = 30000 cm ^ 3
A = 2 (20 * 30) +2 (20 * 50) +2 (30 * 50) = 6200cm ^ 2

b)
V = x ^ 2 * y
A = 2 (x * x) +2 (x * y) +2 (x * y) = 2x ^ 2 +4 xy = 6200 cm ^ 2

I should use the derivatives but I do not know how.

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello nthabiseng p,

Let's first verify the given hint is true. For this I suggest using Lagrange multipliers. I will let the non-negative dimensions of the cuboid be $x,y,z$.

We have the objective function:

\(\displaystyle V(x,y,z)=xyz\)

subject to the constraint:

\(\displaystyle g(x,y,z)=2xy+2xz+2yz-S=0\) where $S$ is the surface area to which we are restricted.

Thus, we find the system:

\(\displaystyle yz=(2y+2z)\lambda\)

\(\displaystyle xz=(2x+2z)\lambda\)

\(\displaystyle xy=(2x+2y)\lambda\)

This system implies that for non-zero dimension values, we must have:

\(\displaystyle x=y=z\)

So we know the given hint is true. Substituting into the constraint, we find:

\(\displaystyle 6x^2=S\,\therefore\,x=\sqrt{\frac{S}{6}}\)

Thus, we should find the dimensions of the cuboid having a given surface area $S$ to be:

\(\displaystyle x=y=z=\sqrt{\frac{S}{6}}\)

Now, assuming we are to use single-variable calculus instead, along with the given hint ($z=x$), we could write:

\(\displaystyle V=x^2y\)

\(\displaystyle S=2x^2+4xy\)

Solving the second equation for $y$, we find:

\(\displaystyle y=\frac{S-2x^2}{4x}\)

Substituting for $y$ into the first equation, there results:

\(\displaystyle V(x)=x^2\left(\frac{S-2x^2}{4x} \right)=\frac{1}{4}\left(Sx-2x^3 \right)\)

Now, differentiating and equating to zero to find the critical point, we get:

\(\displaystyle V'(x)=\frac{1}{4}\left(S-6x^2 \right)=0\)

Since we require \(\displaystyle 0\le x\) we get the critical value:

\(\displaystyle x=\sqrt{\frac{S}{6}}\)

Now, using the second-derivative test, we find:

\(\displaystyle V''(x)=-3x\)

We see the function is concave down for any positive value of $x$, thus we know our critical value is at a maximum, which is global.

And so, using the value for $y(x)$, we find:

\(\displaystyle y=\frac{S-\frac{S}{3}}{4\sqrt{\frac{S}{6}}}=\sqrt{\frac{S}{6}}\)

Hence, we find that:

\(\displaystyle x=y=z=\sqrt{\frac{S}{6}}\)

Using the value \(\displaystyle S=6200\text{ cm}^2\) we then find:

\(\displaystyle x=y=z=\frac{10\sqrt{93}}{3}\,\text{cm}\).