Respuesta :

LammettHash
If you know calculus, you can take the derivative and find the critical point.

The simpler method would be to complete the square. We have

[tex]R=-\dfrac1{800}(x^2-2400x)=-\dfrac1{800}(x^2-2400x+1200^2-1200^2)[/tex]
[tex]R=-\dfrac1{800}\left((x-1200)^2-1200^2\right)=-\dfrac1{800}(x-1200)^2+\dfrac{1200^2}{800}[/tex]

The first term will always be negative, so whenever it vanishes, we're left with the maximum value at [tex]\dfrac{1200^2}{800}=1800[/tex], so the answer is C.

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