Asked by kim
Suppose the revenue (in thousands of dollars) for the sales of 'x' hundred units of an electronic item is given by the function R(x)= 40x^2e^ -0.4x +30. where the maximum capacity of the plant is 800 units. Determine the number of units to produce in order to maximize revenue.
Answers
Answered by
bobpursley
I read you revenue function as
R(x)= 40x<sup>2</sup> *e<sup>-0.4x</sup> +30
dR/dx=40*x^2 *(-.4)e^-.4x+80x*e^(-.4x) =0
or x^2 *( .4)e^-.4x =2 x*e^(-.4x)
x( .4)=2
x=5 hundred units.
check my work.
R(x)= 40x<sup>2</sup> *e<sup>-0.4x</sup> +30
dR/dx=40*x^2 *(-.4)e^-.4x+80x*e^(-.4x) =0
or x^2 *( .4)e^-.4x =2 x*e^(-.4x)
x( .4)=2
x=5 hundred units.
check my work.
Answered by
Damon typo
R(x)= 40x^2 { e^ ???? } -0.4x +30
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