$V=10\pi r^2$

Enter Y1=10πX^2. Use the table to determine where $r$ gives a volume closest to $1000$ cm³. You will find $r\approx 5.6$ cm.

$V=2\pi r^3$

The volume of the tin can is: $500$ cm³. Make a table corresponding to: Y1=2πX^3. You will find $r\approx 4.3$ cm.

$K=200+0.04a$

$I=0.10a$

Use the table to find the first value of $a$ for which $K<I$. You will find $a=3334$

Enter: Y1=250-0.5X^2 with window settings: $0\le x\le 500$ and $0\le y\le 40000$.

Enter: Y1=0.04+200/X with window settings: $0\le x\le 100$ and $0\le y\le 100$.

Enter: Y1=60/(30+0.5X) with window settings: $0\le x\le 500$ and $0\le y\le 3$.

$12.83$ euros

$GTK=\frac{100}{q}+0.1q$

v.a.: $q=0$

When the value of $q$ becomes very large, so does the value of $GTK$ become very large.

$l$ and $b$ represent the length and the width of the printed part of the poster, given in cm. $1$ m² = $10000$ cm². The size of the poster is $(l+25)(b+20)=10000$.

$b+20=\frac{10000}{(l+25)}$ and therefore $b=\frac{10000}{(l+25)}-20$.
GR: Y1=10000/(X+25)-20 with window settings: $0\le x\le 500$ and $0\le y\le 400$.

When $l=0$, it follows that $b=380$ and when $b=0$, it follows that $l=475$. In conclusion $0\le l\le 475$ and $0\le b\le 380$.

In the table you can find that $b=l$ when $l\approx 77.5$ cm.
The dimensions of the poster will become $97.5$ by $102.5$ cm.