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>
The first one is < the second one is < and the last one is =
Area of cone = 1/3 base × height = 1/3 × 8 × 10 = approximately 26.7 m
Ans= approx. 26.4 m
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15x+40=-47+27
15x+40=-20
15x=-60
X=-4
