Which square should be cut to minimize loss?

Question

From a paper size of $950mm × 1200 mm$, squares with a side of $64 mm$ or $46 mm$ can be cut. Which square should be cut to minimize loss?

My attempts:

We have, for square with side 64 mm, the maximum number of squares that can be cut, is equal to $278$

$$950×1200-64^2×278=1312 mm^2$$

For square with side 46 mm, the maximum number of squares that can be cut, is equal to $538$

$$950×1200-46^2×538=1592 mm^2$$

Then, we get $1312<1592$

So, the answer must be $64 mm^2$. Is this answer correct? Because, the answer given in the book is $46 mm$. Where is the error in this approach?

Answer 1

First, you need to find how many of each square can fit the paper: $$\frac{1200}{64}=18.75 \to 18; \frac{950}{64}\approx 14.84\to 14\\ \frac{1200}{46}\approx 26.09\to 26, \frac{950}{46}\approx 20.65\to 20$$ Now calculate the waste (i.e. damage): $$1200\cdot 950-18\cdot 14\cdot 64^2=107,808,\\ 1200\cdot 950-26\cdot 20\cdot 46^2=39,680$$ Hence, the squares $46\times 46$ leave less waste.

Note: It is assumed the sides of squares are parallel to the sides of paper.