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Appendix S1: Computing the optimal realized reproduction
As stated in the text, Williams [18] showed that the optimal level of reproductive effort at a
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given age, E*(x), is that which balances the marginal gain in present reproduction from a
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infinitesimal increase in E against the marginal loss in expected, future reproduction. In a
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continuous model, the optimal value of E is the point where
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dRmax
dRRVmax

dE
dE
.
eq.a1
dRRVmax dRRVmax dRmax

dE
dRmax
dE
,
eq.a2
.
eq.a3
By the chain rule for derivatives,
and combining the two equations above gives
dRRVmax
 1
dRmax
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Thus, to determine the optimal value of E we need to find the point on the curve describing the
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tradeoff between present and future reproduction whose tangent line has slope 1.
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To simplify notation, consider an ellipse with x-intercepts at ±a and y-intercepts at ±b:
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x2 y 2

1
a 2 b2
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We will determine where slope of the tangent line to this curve is precisely dy/dx = -1 for x, y > 0
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dy  x 2 y 2  dy
    1
dx  a 2 b2  dx
,
eq.a5
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2 x 2 y  dy 

 0
a 2 b 2  dx 
,
eq.a6
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2x 2 y

0
a2 b2
,
eq.a7
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b2
y 2 x
a
.
eq.a8
,
eq.a9
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21
.
eq.a4
By substituting this into the ellipse equation and solving for x we get


b4 a 4 x2
x2

1
a2
b2
1
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x2 b2 x 2
 4 1
a2
a
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x2 a 2  b2  a 4

x
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
a2
a 2  b2
,
eq.a10
,
eq.a11
.
eq.a12
.
eq.a13
Replacing a with Rmax(x), b with RRVmax(x), and x with m(x) yields
m( x) 
2
Rmax
( x)
2
2
Rmax
( x)  RRV max
( x)
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