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Let-a-b-c-be-positive-constants-Among-all-real-number-x-and-y-satisfying-ax-by-c-find-the-maximum-value-of-product-xy-




Question Number 138797 by bramlexs22 last updated on 18/Apr/21
Let a,b,c be positive constants.  Among all real number x and y   satisfying ax+by=c , find the  maximum value of product  xy.
$${Let}\:{a},{b},{c}\:{be}\:{positive}\:{constants}. \\ $$$${Among}\:{all}\:{real}\:{number}\:{x}\:{and}\:{y}\: \\ $$$${satisfying}\:{ax}+{by}={c}\:,\:{find}\:{the} \\ $$$${maximum}\:{value}\:{of}\:{product} \\ $$$${xy}. \\ $$
Answered by TheSupreme last updated on 18/Apr/21
if b=0 or a=0 xy is unlimited  else: y=−(a/b)x−(c/b)  (a/b)=m   (c/b)=q  y=−mx−q  σ=xy=x(mx+q)=−mx^2 −qx=−x(mx+q)  (∂σ/∂x)=−2mx−q=0 →x=−(q/(2m))  σ_(max) =(q/(2m))((q/2))=(q^2 /(4m))=(c^2 /(4ab))
$${if}\:{b}=\mathrm{0}\:{or}\:{a}=\mathrm{0}\:{xy}\:{is}\:{unlimited} \\ $$$${else}:\:{y}=−\frac{{a}}{{b}}{x}−\frac{{c}}{{b}} \\ $$$$\frac{{a}}{{b}}={m}\: \\ $$$$\frac{{c}}{{b}}={q} \\ $$$${y}=−{mx}−{q} \\ $$$$\sigma={xy}={x}\left({mx}+{q}\right)=−{mx}^{\mathrm{2}} −{qx}=−{x}\left({mx}+{q}\right) \\ $$$$\frac{\partial\sigma}{\partial{x}}=−\mathrm{2}{mx}−{q}=\mathrm{0}\:\rightarrow{x}=−\frac{{q}}{\mathrm{2}{m}} \\ $$$$\sigma_{{max}} =\frac{{q}}{\mathrm{2}{m}}\left(\frac{{q}}{\mathrm{2}}\right)=\frac{{q}^{\mathrm{2}} }{\mathrm{4}{m}}=\frac{{c}^{\mathrm{2}} }{\mathrm{4}{ab}} \\ $$
Answered by EDWIN88 last updated on 18/Apr/21
without calculus.  let  { ((ax = u)),((by = v)) :} ⇒ u+v = c ; so u.v minimum  where  { ((u=(c/2))),((v=(c/2))) :} then (uv)_(min) = (c^2 /4)  ⇒(ax.by)_(min) = (c^2 /4) or xy_(min)  = (c^2 /(4ab)).
$$\mathrm{without}\:\mathrm{calculus}. \\ $$$$\mathrm{let}\:\begin{cases}{\mathrm{ax}\:=\:\mathrm{u}}\\{\mathrm{by}\:=\:\mathrm{v}}\end{cases}\:\Rightarrow\:\mathrm{u}+\mathrm{v}\:=\:\mathrm{c}\:;\:\mathrm{so}\:\mathrm{u}.\mathrm{v}\:\mathrm{minimum} \\ $$$$\mathrm{where}\:\begin{cases}{\mathrm{u}=\frac{\mathrm{c}}{\mathrm{2}}}\\{\mathrm{v}=\frac{\mathrm{c}}{\mathrm{2}}}\end{cases}\:\mathrm{then}\:\left(\mathrm{uv}\right)_{\mathrm{min}} =\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\left(\mathrm{ax}.\mathrm{by}\right)_{\mathrm{min}} =\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{or}\:\mathrm{xy}_{\mathrm{min}} \:=\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{4ab}}.\: \\ $$

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