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# Logarithm Question (Important)

3 years, 8 months ago

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$$P=2X^2+(3-X)^2$$) Minimum of P is at $$X=1$$ or $$(x,y)=(2,4)$$ which is $$6$$ · 3 years, 8 months ago

I thought the minimum of P is 6 because from the equation P=2X^2 + (3-X)^2=3X^2-6X+9 by completing the square P=3(X-1)^2 + 6, hence the minimum value is 6. Even if we consider your answer that 0 is minimum then log x base 2 has to be zero > then x=1 and repeating the same argument we find that y=1 and unfortunately this contradicts the statement that xy=8 because if minimum P is 0 then xy has to be 1. · 3 years, 8 months ago

You are right. · 3 years, 8 months ago

from what P=2X^2+(3−X)^2 ? · 3 years, 8 months ago

The questions asks us to express P in terms of X. · 3 years, 8 months ago

i understood about 2x^2, but i have question, from what (3−X)^2 ? · 3 years, 8 months ago