Answer:
3
Step-by-step explanation:
The surface area of a cylindrical can is equal to the sum of the area of two circles and the body of the cylinder: 2πr2 + 2πrh. volume is equal to π<span>r2h.
V = </span>π<span>r2h = 128 pi
r2h = 128
h = 128/r2
A = </span><span>2πr2 + 2πrh
</span>A = 2πr2 + 2πr*(<span>128/r2)
</span>A = 2πr2 + 256 <span>π / r
</span><span>
the optimum dimensions is determined by taking the first derivative and equating to zero.
dA = 4 </span>πr - 256 <span>π /r2 = 0
r = 4 cm
h = 8 cm
</span><span>
</span>
Answer:
y=.44-.66
Step-by-step explanation:
hope this helps not 100 percent sure let me know in the comments.
Lengths are given to 1dp so the biggest that the lengths could be is 4.65cm and 5.35cm. if you use pythagoras on these to find hypotenuse, this is the upper bound.
The first one seems more reasonable
