Find-minimum-value-of-2x-2-2xy-4y-5y-2-x-for-x-and-y-real-numbers-

Question Number 188806 by horsebrand11 last updated on 07/Mar/23

$${Find}\:{minimum}\:{value}\:{of}\: \\ $$$$\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{4}{y}+\mathrm{5}{y}^{\mathrm{2}} −{x}\: \\ $$$$\:{for}\:{x}\:{and}\:{y}\:{real}\:{numbers} \\ $$

Answered by mr W last updated on 07/Mar/23

Method I k=2x^2 +2xy+4y+5y^2 −x 2x^2 +(2y−1)x+5y^2 +4y−k=0 Δ=(2y−1)^2 −8(5y^2 +4y−k)≥0 −36y^2 −36y+8k+1≥0 36y^2 +36y−8k−1≤0 Δ=36^2 +4×36(8k+1)≥0 ⇒k≥−(5/4) (2x^2 +2xy+4y+5y^2 −x)_(min) =−(5/4)

$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$${k}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{4}{y}+\mathrm{5}{y}^{\mathrm{2}} −{x} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\left(\mathrm{2}{y}−\mathrm{1}\right){x}+\mathrm{5}{y}^{\mathrm{2}} +\mathrm{4}{y}−{k}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}{y}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{5}{y}^{\mathrm{2}} +\mathrm{4}{y}−{k}\right)\geqslant\mathrm{0} \\ $$$$−\mathrm{36}{y}^{\mathrm{2}} −\mathrm{36}{y}+\mathrm{8}{k}+\mathrm{1}\geqslant\mathrm{0} \\ $$$$\mathrm{36}{y}^{\mathrm{2}} +\mathrm{36}{y}−\mathrm{8}{k}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\Delta=\mathrm{36}^{\mathrm{2}} +\mathrm{4}×\mathrm{36}\left(\mathrm{8}{k}+\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{k}\geqslant−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{4}{y}+\mathrm{5}{y}^{\mathrm{2}} −{x}\right)_{{min}} =−\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Answered by mr W last updated on 07/Mar/23

Method II k=2x^2 +2xy+4y+5y^2 −x (∂k/∂x)=4x+2y−1=0 ...(i) (∂k/∂y)=2x+4+10y=0 ...(ii) ⇒x=(1/2), y=−(1/2) k_(min) =2×(1/4)−(1/2)−2+(5/4)−(1/2)=−(5/4) ✓ (it′s minimum, because (∂^2 k/∂x^2 )>0, (∂^2 k/∂y^2 )>0)

$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$${k}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{4}{y}+\mathrm{5}{y}^{\mathrm{2}} −{x} \\ $$$$\frac{\partial{k}}{\partial{x}}=\mathrm{4}{x}+\mathrm{2}{y}−\mathrm{1}=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\frac{\partial{k}}{\partial{y}}=\mathrm{2}{x}+\mathrm{4}+\mathrm{10}{y}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}},\:{y}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${k}_{{min}} =\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}+\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{5}}{\mathrm{4}}\:\checkmark \\ $$$$\left({it}'{s}\:{minimum},\:{because}\:\frac{\partial^{\mathrm{2}} {k}}{\partial{x}^{\mathrm{2}} }>\mathrm{0},\:\frac{\partial^{\mathrm{2}} {k}}{\partial{y}^{\mathrm{2}} }>\mathrm{0}\right) \\ $$

Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23

Sir can you check my reply on your comment in Q188362

Commented by mr W last updated on 07/Mar/23

$${yes},\:{you}\:{got}\:{the}\:{right}\:{result}!\:{thanks}! \\ $$$${sorry},\:{i}\:{don}'{t}\:{get}\:{notified}\:{about}\:{your} \\ $$$${reply}. \\ $$

Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23

OK Sir. Me too. I don′t get notified about replies.

$$\mathrm{OK}\:\mathrm{Sir}. \\ $$$$\mathrm{Me}\:\mathrm{too}.\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{notified}\:\mathrm{about}\:\mathrm{replies}. \\ $$

Commented by mr W last updated on 08/Mar/23

can you check my answer in Q111118 with a program? thanks sir!

$${can}\:{you}\:{check}\:{my}\:{answer}\:{in}\:{Q}\mathrm{111118} \\ $$$${with}\:{a}\:{program}?\:{thanks}\:{sir}! \\ $$

Commented by ARUNG_Brandon_MBU last updated on 09/Mar/23

The system says "Question deleted from forum."

Commented by mr W last updated on 09/Mar/23

sorry the question # is 188879. i mistook the ID of that user for the question number.

$${sorry}\:{the}\:{question}\:#\:{is}\:\mathrm{188879}. \\ $$$${i}\:{mistook}\:{the}\:{ID}\:{of}\:{that}\:{user}\:{for}\:{the} \\ $$$${question}\:{number}. \\ $$

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