Answer to Question #107942 in Calculus for annie

Question #107942
Use Lagrange multipliers to find the point (a,b) on the graph of y=e^8x, where the value ab is as small as possible.

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Expert's answer
2020-04-10T17:34:08-0400

Consider the objective function "f(x,y)=xy"

Let the constraint function be "g(x,y)=y\u2212e^{8x}"


Use Lagrange Multiplier as follows:


"\\nabla f(x,y)=\\lambda \\nabla g(x,y)"

Next, find the gradients both sides as,


"<y,x>=\\lambda <-8e^{8x},1>"

Equate both sides to obtain,


"y=\u22128\u03bbe^{8x}"


"x=\u03bb"

Substitute "x" for "\\lambda" into relation "y=\u22128\u03bbe^{8x}" to obtain "y=\u22128xe^{8x}"

Now, substitute "\u22128xe^{8x}" for "y" into constraint equation "y\u2212e^{8x}=0" and solve for "x" as,


"\u22128xe^{8x}\u2212e^{8x}=0"


"e^{8x}(\u22128x\u22121)=0"


"\u22128x\u22121=0,e^{8x}\\ne0"


"x=-\\frac{1}{8}"


Plug "x=-\\frac{1}{8}" into relation "y=\u22128xe^{8x}" and solve for "y" as,


"y=\u22128(-\\frac{1}{8})e^{8(-\\frac{1}{8})}=e^{-1}"


Therefore, the required point "(a,b)" on the graph of "y=e^{8x}" is "(-\\frac{1}{8},\\frac{1}{e})"

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