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max of sine is 1, min is -1 replace \(\sin(\frac{\pi t}{5})\) by 1 and see what you get
\[20 \sin (\pi t/5) + 12 \]\[20 \overbrace{\sin (\pi t/5)}^{\text{ put a 1 here}} + 12 \]
then \[20 \overbrace{\sin (\pi t/5)}^{\text{ put a -1 here}} + 12\]
20 + 12 = 32 for the maximum?
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no
\(-20+8\)
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