Answer:
73%
Step-by-step explanation:
If the diameter (d) and height (h) of the small can is increased by 20%, then the diameter (D) of the large can will be 1.2d and the height of the large can (H) will be equal to 1.2h.
Now, the volume of the small can, [tex]v = \pi (\frac{d}{2}) ^{2} h = \frac{\pi d^{2} h }{4}[/tex].
Now, the volume of the large can will be [tex]V = \pi (\frac{D}{2} )^{2} H = \frac{\pi D^{2} H }{4} = \frac{\pi (1.2d)^{2} \times (1.2h) }{4} = 1.728 \times \frac{\pi d^{2} h }{4}[/tex]
Therefore, the percentage increase in volume from smaller can to larger can will be [tex]\frac{1.728 \times \frac{\pi d^{2} h }{4} - \frac{\pi d^{2} h }{4}}{\frac{\pi d^{2} h }{4}} \times 100[/tex]
= 72.8%
≈ 73% (Answer)