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I have a 3D rectangular object. It has a volume of $75\text{cm}^3$; it has a square base with one edge of the base having a length of $x$. Find a function $S$, that represents the surface area of the object with respect to $x$.

I know that the surface area $S$ of a 3D rectangular object is a function of its length ($L$), width ($W$), and height ($H$). That is, adding the areas of each face gives the surface area. The area of each face is below:

$$\begin{align} S_1 &= LW \\ S_2 &= LW \\ S_3 &= WH \\ S_4 &= WH \\ S_5 &= HL \\ S_6 &= HL \end{align}$$

Or more declarative, the surface area of a 3D rectangular object is:

$$\begin{align} S &= LW + LW + WH + WH + HL + HL \\ &= 2LW + 2WH + 2HL \end{align}$$

I also know the volume $V$ of a 3D rectangular object is a function of its length, width, and height:

$$V = LWH$$

Using algebra I can find any given dimension:

$$\begin{align} L &= V/WH \\ W &= V/LH \\ H &= V/LW \end{align}$$

Going back to the shape at hand, I know the area of one of the faces of the polygon: $x^2$. This corresponds to the length multiplied by width.

Since I know the volume, length, and width, I can get the height, too:

$$\begin{align} H &= V/LW \\ &= 75/x^2 \end{align}$$

Substituting $H$ into the equation for the surface area of a 3D rectangular object:

$$\begin{align} S &= 2LW + 2WH + 2HL \\ &= 2(x^2) + 2(\frac{75}{x^2})(x) + 2(\frac{75}{x^2})(x) \\ &= 2x^2 + 2x(\frac{75}{x^2}) + 2x(\frac{75}{x^2}) \\ &= 2x^2 + 4x(\frac{75}{x^2}) \\ &= 2x^2 + \frac{4x \cdot 75}{x^2} \\ &= 2x^2 + \frac{300x}{x^2} \\ &= 2x^2 + 300 \cdot \frac{x}{x^2} \end{align}$$

However, this has been marked incorrect. What's the issue with my process?

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It is the volume of the object that is $LWH$ and is given in cm$^3$. If you mean the volume is $75$ cm$^3$ the surface area is $2x^2+4xH$, which is $2x^2+\frac {300}x$ as you say. Maybe they expected you to replace $\frac x{x^2}$ with $\frac 1x$

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  • $\begingroup$ I've corrected it: gets confusing dealing with 3D and 2D interchangeably in one question. $\endgroup$ – gator Aug 30 '18 at 20:38

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