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If x is real then find the maximum value of (3x^2+9x+17)/(3x^2+9x+7)3x2+9x+173x2+9x+7.

2 Answers
P dilip_k
Jun 29, 2018

Let

y=(3x^2+9x+17)/(3x^2+9x+7)y=3x2+9x+173x2+9x+7

=>y(3x^2+9x+7)=(3x^2+9x+17)⇒y(3x2+9x+7)=(3x2+9x+17)

=>3(y-1)x^2+9(y-1)x+(7y-17)=0⇒3(y−1)x2+9(y−1)x+(7y−17)=0

As xx is real we can write

(9(y-1))^2 -4*3(y-1)(7y-17)>=0(9(y−1))2−4⋅3(y−1)(7y−17)≥0

=>81(y-1)^2 -4*3(y-1)(7y-17)>=0⇒81(y−1)2−4⋅3(y−1)(7y−17)≥0
=>27(y-1)^2 -4(y-1)(7y-17)>=0⇒27(y−1)2−4(y−1)(7y−17)≥0

=>27y^2-54y+27 -4(7y^2-24y+17)>=0⇒27y2−54y+27−4(7y2−24y+17)≥0

=>27y^2-54y+27 -28y^2+96y-68>=0⇒27y2−54y+27−28y2+96y−68≥0

=>-y^2+42y-41>=0⇒−y2+42y−41≥0

=>y^2-42y+41<=0⇒y2−42y+41≤0

=>y^2-41y-y+41<=0⇒y2−41y−y+41≤0

=>y(y-41)-1(y-41)<=0⇒y(y−41)−1(y−41)≤0

=>(y-41)(y-1)<=0⇒(y−41)(y−1)≤0

This inequality holds for

1 <=y<=411≤y≤41

Hence maximum value of the given expression for real values of x will be 4141

Ananda Dasgupta
Jun 29, 2018

41

Explanation:

(3x^2+9x+17)/(3x^2+9x+7)=1+10/(3x^2+9x+7)3x2+9x+173x2+9x+7=1+103x2+9x+7

Hence, to maximize (3x^2+9x+17)/(3x^2+9x+7)3x2+9x+173x2+9x+7 we need to minimize 3x^2+9x+73x2+9x+7. Now

3x^2+9x+7 = 3(x^2+3x)+73x2+9x+7=3(x2+3x)+7
qquad= 3(x^2+2*x*3/2+(3/2)^2)+7-3(3/2)^2
qquad = 2(x+3/2)^2+1/4

Since (x+3/2)^2 >= 0 we have

3x^2+9x+7>= 1/4 implies
10/(3x^2+9x+7) <= 40 implies

(3x^2+9x+17)/(3x^2+9x+7)=1+10/(3x^2+9x+7)<= 41

Thus, the maximum value is 41 (attained for x=-3/2).

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