Menu Close

Let-m-n-be-given-positive-integers-If-x-amp-y-are-positive-numbers-such-that-x-y-S-S-is-a-constant-find-the-value-of-x-and-y-that-maximize-Q-x-m-y-n-

Tinku Tara 0 Comments
Pin




Question Number 140734 by liberty last updated on 12/May/21
Let m,n be given positive integers. If x & y are positive numbers such that x+y= S , S is a constant, find the value of x and y that maximize Q=x^m y^n .
$$\mathrm{Let}\:\mathrm{m},\mathrm{n}\:\mathrm{be}\:\mathrm{given}\:\mathrm{positive}\:\mathrm{integers}. \\ $$$$\mathrm{If}\:\mathrm{x}\:\&\:\mathrm{y}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{x}+\mathrm{y}=\:\mathrm{S}\:,\:\mathrm{S}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{that}\:\mathrm{maximize} \\ $$$$\mathrm{Q}=\mathrm{x}^{\mathrm{m}} \mathrm{y}^{\mathrm{n}} \:. \\ $$
Commented by mr W last updated on 12/May/21
if m and n both are even, there is minimum. if only one of m, n is even, there is no maximum and no minimum. if both m and n are odd, there is maximum.
$${if}\:{m}\:{and}\:{n}\:{both}\:{are}\:{even},\:{there}\:{is}\: \\ $$$${minimum}. \\ $$$$ \\ $$$${if}\:{only}\:{one}\:{of}\:{m},\:{n}\:{is}\:{even},\:{there}\:{is}\:{no} \\ $$$${maximum}\:{and}\:{no}\:{minimum}. \\ $$$$ \\ $$$${if}\:{both}\:{m}\:{and}\:{n}\:{are}\:{odd},\:{there}\:{is}\: \\ $$$${maximum}. \\ $$
Answered by EDWIN88 last updated on 12/May/21
setting (dQ/dx) = mx^(m−1) (S−x)^n −n(S−x)^(n−1) x^m = 0 we obtain x = ((mS)/(m+n)) . The first−derivative test show that this yields a relative maximum and therefore by the uniqueness of critical number, an absolute maximum when x=((mS)/(m+n)) , y=((nS)/(m+n))
$$\mathrm{setting}\:\frac{\mathrm{dQ}}{\mathrm{dx}}\:=\:\mathrm{mx}^{\mathrm{m}−\mathrm{1}} \left(\mathrm{S}−\mathrm{x}\right)^{\mathrm{n}} −\mathrm{n}\left(\mathrm{S}−\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{m}} \:=\:\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{obtain}\:\mathrm{x}\:=\:\frac{\mathrm{mS}}{\mathrm{m}+\mathrm{n}}\:.\:\mathrm{The}\:\mathrm{first}−\mathrm{derivative}\:\mathrm{test} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{yields}\:\mathrm{a}\:\mathrm{relative}\:\mathrm{maximum}\: \\ $$$$\mathrm{and}\:\mathrm{therefore}\:\mathrm{by}\:\mathrm{the}\:\mathrm{uniqueness}\:\mathrm{of}\:\mathrm{critical} \\ $$$$\mathrm{number},\:\mathrm{an}\:\mathrm{absolute}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{x}=\frac{\mathrm{mS}}{\mathrm{m}+\mathrm{n}}\:,\:\mathrm{y}=\frac{\mathrm{nS}}{\mathrm{m}+\mathrm{n}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *