a b c a \ b \ c , 1 2 3 1 \ 2 \ 3

Calculus Level pending

Find the value of a \displaystyle a so that f ( x ) = a x e b x 2 \displaystyle f(x) = axe^{bx^2} has a maximum value of f ( 3 ) = 1 \displaystyle f(3) = 1 .

e 3 \displaystyle \frac{\sqrt{e}}{3} 3 5 \displaystyle \frac{3}{\sqrt{5}} 3 e \displaystyle \frac{3}{\sqrt{e}} 1 3 e 9 \displaystyle \frac{1}{3e^9} e 5 3 \displaystyle \frac{e^{\sqrt{5}}}{3}

Zach Abueg by Zach Abueg , 22, USA

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1 solution

Zach Abueg
Apr 17, 2017

f ( 3 ) = 3 a e 9 b = 1 \displaystyle f(3) = 3ae^{9b} = 1

f ′ ( x ) = ( a x ) ( e b x 2 ) ( 2 b x ) + a e b x 2 = a e b x 2 ( 2 b x 2 + 1 ) \displaystyle f'(x) = (ax)(e^{bx^2})(2bx) + ae^{bx^2} = ae^{bx^2}(2bx^2 + 1)

f ′ ( 3 ) = a e 9 b ( 18 b + 1 ) = 0 \displaystyle f'(3) = ae^{9b}(18b + 1) = 0

18 b + 1 = 0 ⟹ b = − 1 18 \displaystyle 18b + 1 = 0 \Longrightarrow b = - \frac{1}{18}

f ( 3 ) = 3 a e 9 × − 1 18 = 3 a e − 1 2 = 1 \displaystyle f(3) = 3ae^{9 \ \times \ - \frac{1}{18}} = 3ae^{- \frac 12} = 1

a = 1 3 e − 1 2 = e 3 \displaystyle a = \frac {1}{3e^{- \frac 12}} = \frac {\sqrt{e}}{3}

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