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

AI Thread Summary
The discussion focuses on maximizing the volume of a cuboid with a fixed surface area. The initial box dimensions are given, and the volume and surface area are calculated as 30,000 cm³ and 6,200 cm², respectively. To find the dimensions of a new box with the same surface area that maximizes volume, it is suggested to use derivatives and Lagrange multipliers. The conclusion indicates that the optimal dimensions for maximum volume occur when all sides are equal, leading to a cubic shape where each side measures approximately 10√93/3 cm. This approach confirms that the dimensions yielding the largest volume under the given constraints are equal.
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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.
 
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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:

$$V(x,y,z)=xyz$$

subject to the constraint:

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

Thus, we find the system:

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

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

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

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

$$x=y=z$$

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

$$6x^2=S\,\therefore\,x=\sqrt{\frac{S}{6}}$$

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

$$x=y=z=\sqrt{\frac{S}{6}}$$

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

$$V=x^2y$$

$$S=2x^2+4xy$$

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

$$y=\frac{S-2x^2}{4x}$$

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

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

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

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

Since we require $$0\le x$$ we get the critical value:

$$x=\sqrt{\frac{S}{6}}$$

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

$$V''(x)=-3x$$

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:

$$y=\frac{S-\frac{S}{3}}{4\sqrt{\frac{S}{6}}}=\sqrt{\frac{S}{6}}$$

Hence, we find that:

$$x=y=z=\sqrt{\frac{S}{6}}$$

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

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