1) We note here that each specific divisor x (3, 4, 5, 6 or 7) of the desired integer N yields a remainder of (x - 2). Therefore, each specified divisor x of N + 2 yields a remainder of 0. In other words, N +2 is a multiple of 3, 4, 5, 6 and 7. The smallest such number is:
420 (2 · 2 · 3 · 5 · 7)
If N + 2 = 420 then the desired number N is: N = 420 - 2 = 418
Following on from this:
418/ 3 = 139 r1
418/4 = 104 r2
418/5 = 83 r3
418/6 = 69 r4
418/7 = 59 r5
2) Write the formula for the volume of a cone:
V = 1/3 (pr2 h)
Also, for this situation: r2 + h 2 = 1: V= 1/3 pr2 Ö(1 - r2)
We see the volume V is zero when r = 0 or 1. Hence, the maximum ratio must be found for some r between 0 and 1. How to obtain it? We first introduce the formula for the surface area of a cone:
A = pr h s where h s denotes the slant height of the cone for which:
h s = Ö (r2 + h 2)
Now, because in our case: h s = 1 then A = pr.
Then the ratio V/ A = 1/3 pr2 Ö(1 - r2)/ pr
Or, more simply:
V/ A = 1/3 r Ö(1 - r2)
Differentiation of a function must be used at this point. The maximum or minimum of a differentiable function occurs at critical points where the derivative is zero. Then let (V/A) be a differentiable function, e.g.
(V/ A)' = 1/3 r Ö(1 - r2)
By the product rule:
1/3 r Ö(1 - r2) = 1/3 Ö(1 - r2) + 1/3 r Ö(1 - r2)
Using the chain rule for derivatives:
(V/ A)' = 1/3 Ö(1 - r2) + 1/3 r (½) Ö(1 - r2) (2r)
= 1/3 Ö(1 - r2) - r2/3 Ö(1 - r2)
We then set the derivative to 0 and solve for r:
1/3 Ö(1 - r2) - r2/3 Ö(1 - r2) = 0
Multiply both sides by: 3 Ö(1 - r2):
0= (1 - r2) - r2 = 1 - 2r2
Or: 2r2 = 1
è
r = Ö1/2 = Ö2/2
Then Max (V/A) = (1/3) (Ö2/2) (Ö2/2) = 2/ 12 = 1/6
Thus the maximum ratio (V/A) occurs when:
r = Ö2/2 and h = Ö2/2
This can be achieved when one full quadrant of a circle (sector angle 90 deg) is removed from a paper circle and the straight edges as re-connected.
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