標題:
一條很難的數學問題,求解答
f(x, y) = 3-x^2+6x-y^2+4y is the elevation at location (x,y). As I walk in a straight line from point A = (0,2) to point B=(3,0) what is my maximum elevation along the path?
最佳解答:
f(x，y)=3－x^2+6x－y^2+4y is the elevation at location(x，y). As I walkin a straight line from point A=(0，2) to point B=(3，0) what is my maximum elevation along the path? Sol AB：(0－2)/(3－0)=(y－2)/(x－0) －2x=3y－6 2x+3y=6 3y=6－2x g(x)=3－x^2+6x－[6－2x)/3]^2+4*(6－2x)/3 9g(x)=27－9x^2+54x－(4x^2－24x+36)+72－24x =－13x^2+54x+63 =－13(x^2－54x/13)+63 =－13[x^2－2*x*(27/13)+(27/13)^2]+63+729/13 =－13(x－27/13)^2+63+729/13 0<=x<=3 －27/13<=x－27/13<=12/13 0<=(x－27/13)^2<=729/169 0<=13(x－27/13)^2<=729/13 －729/13<=－13(x－27/13)^2<=0 63+729/13－729/13<=－13(x-27/13)^2+63+729/13<=63+729/13 63<=9g(x)<=1548/13 7<=g(y)<=1548/117 7<=f(x,y)<=1548/117
其他解答:
一條很難的數學問題,求解答
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發問:f(x, y) = 3-x^2+6x-y^2+4y is the elevation at location (x,y). As I walk in a straight line from point A = (0,2) to point B=(3,0) what is my maximum elevation along the path?
最佳解答:
f(x，y)=3－x^2+6x－y^2+4y is the elevation at location(x，y). As I walkin a straight line from point A=(0，2) to point B=(3，0) what is my maximum elevation along the path? Sol AB：(0－2)/(3－0)=(y－2)/(x－0) －2x=3y－6 2x+3y=6 3y=6－2x g(x)=3－x^2+6x－[6－2x)/3]^2+4*(6－2x)/3 9g(x)=27－9x^2+54x－(4x^2－24x+36)+72－24x =－13x^2+54x+63 =－13(x^2－54x/13)+63 =－13[x^2－2*x*(27/13)+(27/13)^2]+63+729/13 =－13(x－27/13)^2+63+729/13 0<=x<=3 －27/13<=x－27/13<=12/13 0<=(x－27/13)^2<=729/169 0<=13(x－27/13)^2<=729/13 －729/13<=－13(x－27/13)^2<=0 63+729/13－729/13<=－13(x-27/13)^2+63+729/13<=63+729/13 63<=9g(x)<=1548/13 7<=g(y)<=1548/117 7<=f(x,y)<=1548/117
其他解答:
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