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Question Number 110669 by Aina Samuel Temidayo last updated on 30/Aug/20
Given that f(x)=(3+2x)^3 (4−x)^4 on the interval −(3/2)<x<4. Find the (a) Maximum value of f(x) (b) The value of x that gives the maximum in (a)
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{3}+\mathrm{2x}\right)^{\mathrm{3}} \left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{4}} \:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{interval}\:−\frac{\mathrm{3}}{\mathrm{2}}<\mathrm{x}<\mathrm{4}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{that}\:\mathrm{gives}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{in}\:\left(\mathrm{a}\right) \\ $$
Commented by Her_Majesty last updated on 30/Aug/20
(3+2x)^3 (4−x^4 ) or rather (3+2x)^3 (4−x)^4 ?
$$\left(\mathrm{3}+\mathrm{2}{x}\right)^{\mathrm{3}} \left(\mathrm{4}−{x}^{\mathrm{4}} \right)\:{or}\:{rather}\:\left(\mathrm{3}+\mathrm{2}{x}\right)^{\mathrm{3}} \left(\mathrm{4}−{x}\right)^{\mathrm{4}} ? \\ $$
Commented by Aina Samuel Temidayo last updated on 30/Aug/20
You are correct. I have changed it. Thanks.
$$\mathrm{You}\:\mathrm{are}\:\mathrm{correct}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{changed}\:\mathrm{it}. \\ $$$$\mathrm{Thanks}. \\ $$
Answered by Her_Majesty last updated on 30/Aug/20
f′(x)=6(3+2x)^2 (4−x)^4 −4(4−x)^3 (3+2x)^3 = =2(3+2x)^2 (4−x)^3 (3(4−x)−2(3+2x))= =2(3+2x)^2 (4−x)^3 (6−7x) f′(x)=0 ⇒ x_(1,2) =−(3/2)∧x_(3,4,5) =4∧x_6 =(6/7) now obviously f(−(3/2))=f(4)=0 but f((6/7))=((2^4 3^3 11^7 )/7^7 )=((8418457872)/(823543))≈10222.2
$${f}'\left({x}\right)=\mathrm{6}\left(\mathrm{3}+\mathrm{2}{x}\right)^{\mathrm{2}} \left(\mathrm{4}−{x}\right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{4}−{x}\right)^{\mathrm{3}} \left(\mathrm{3}+\mathrm{2}{x}\right)^{\mathrm{3}} = \\ $$$$=\mathrm{2}\left(\mathrm{3}+\mathrm{2}{x}\right)^{\mathrm{2}} \left(\mathrm{4}−{x}\right)^{\mathrm{3}} \left(\mathrm{3}\left(\mathrm{4}−{x}\right)−\mathrm{2}\left(\mathrm{3}+\mathrm{2}{x}\right)\right)= \\ $$$$=\mathrm{2}\left(\mathrm{3}+\mathrm{2}{x}\right)^{\mathrm{2}} \left(\mathrm{4}−{x}\right)^{\mathrm{3}} \left(\mathrm{6}−\mathrm{7}{x}\right) \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1},\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}\wedge{x}_{\mathrm{3},\mathrm{4},\mathrm{5}} =\mathrm{4}\wedge{x}_{\mathrm{6}} =\frac{\mathrm{6}}{\mathrm{7}} \\ $$$${now}\:{obviously}\:{f}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)={f}\left(\mathrm{4}\right)=\mathrm{0}\:{but} \\ $$$${f}\left(\frac{\mathrm{6}}{\mathrm{7}}\right)=\frac{\mathrm{2}^{\mathrm{4}} \mathrm{3}^{\mathrm{3}} \mathrm{11}^{\mathrm{7}} }{\mathrm{7}^{\mathrm{7}} }=\frac{\mathrm{8418457872}}{\mathrm{823543}}\approx\mathrm{10222}.\mathrm{2} \\ $$
Answered by 1549442205PVT last updated on 30/Aug/20
Apply Cauchy′s inequality we have (3+2x)^3 (4−x)^4 =(((3+2x)^3 (6−1.5x)^4 )/(1.5^4 ))(1) Apply Cauchy′s inequality we have 7^7 (√((3+2x)(3+2x)(3+2x)(6−1.5x)(6−1.5x)(7−1.5x)(6−1.5x))) ≤(3+2x)+(3+2x)+(3+2x)+(6−1.5x)+(6−1.5x)+(7−1.5x)+(6−1.5x) =33 (2) ⇒(3+2x)^3 (6−1.5x)^4 ≤(((33)/7))^7 (((3+2x)^3 (6−1.5x)^4 )/(1.5^4 ))≤((33^7 )/(7^7 .1.5^4 )) The equality ocurrs if and only if 3+2x=6−1.5x⇔3.5x=3⇔x=(6/7) Thus f(x) has greatest value equal to ((33^7 )/(7^7 .1.5^4 )) when x=(6/7)
$$ \\ $$$$\mathrm{Apply}\:\mathrm{Cauchy}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{3}+\mathrm{2x}\right)^{\mathrm{3}} \left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{4}} =\frac{\left(\mathrm{3}+\mathrm{2x}\right)^{\mathrm{3}} \left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)^{\mathrm{4}} }{\mathrm{1}.\mathrm{5}^{\mathrm{4}} }\left(\mathrm{1}\right) \\ $$$$\mathrm{Apply}\:\mathrm{Cauchy}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{7}\:^{\mathrm{7}} \sqrt{\left(\mathrm{3}+\mathrm{2x}\right)\left(\mathrm{3}+\mathrm{2x}\right)\left(\mathrm{3}+\mathrm{2x}\right)\left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)\left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)\left(\mathrm{7}−\mathrm{1}.\mathrm{5x}\right)\left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)} \\ $$$$\leqslant\left(\mathrm{3}+\mathrm{2x}\right)+\left(\mathrm{3}+\mathrm{2x}\right)+\left(\mathrm{3}+\mathrm{2x}\right)+\left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)+\left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)+\left(\mathrm{7}−\mathrm{1}.\mathrm{5x}\right)+\left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right) \\ $$$$=\mathrm{33}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\left(\mathrm{3}+\mathrm{2x}\right)^{\mathrm{3}} \left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)^{\mathrm{4}} \leqslant\left(\frac{\mathrm{33}}{\mathrm{7}}\right)^{\mathrm{7}} \\ $$$$\frac{\left(\mathrm{3}+\mathrm{2x}\right)^{\mathrm{3}} \left(\mathrm{6}−\mathrm{1}.\mathrm{5x}\right)^{\mathrm{4}} }{\mathrm{1}.\mathrm{5}^{\mathrm{4}} }\leqslant\frac{\mathrm{33}^{\mathrm{7}} }{\mathrm{7}^{\mathrm{7}} .\mathrm{1}.\mathrm{5}^{\mathrm{4}} } \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\: \\ $$$$\mathrm{3}+\mathrm{2x}=\mathrm{6}−\mathrm{1}.\mathrm{5x}\Leftrightarrow\mathrm{3}.\mathrm{5x}=\mathrm{3}\Leftrightarrow\mathrm{x}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\mathrm{Thus}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{greatest}\:\mathrm{value}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\frac{\mathrm{33}^{\mathrm{7}} }{\mathrm{7}^{\mathrm{7}} .\mathrm{1}.\mathrm{5}^{\mathrm{4}} }\:\mathrm{when}\:\mathrm{x}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$
Commented by Aina Samuel Temidayo last updated on 30/Aug/20
Yes. Both are correct.
$$\mathrm{Yes}.\:\mathrm{Both}\:\mathrm{are}\:\mathrm{correct}. \\ $$

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