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If-x-3-2-y-4-2-25-what-maximum-and-minimum-value-of-7x-8y-

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Question Number 140270 by liberty last updated on 06/May/21
If (x−3)^2 +(y−4)^2 = 25 , what maximum and minimum value of 7x+8y ?
$$\mathrm{If}\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{25}\:, \\ $$$$\mathrm{what}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{7x}+\mathrm{8y}\:? \\ $$
Answered by EDWIN88 last updated on 06/May/21
let { ((x−3=5cos θ→x=3+5cos θ)),((y−4=5sin θ→y=4+5sin θ)) :} let 7(3+5cos θ)+8(4+5sin θ)=f(θ) f(θ)= 35cos θ+40sin θ+53 f(θ)=(√(35^2 +40^2 )) cos (θ−α)+53 where α = tan^(−1) ((8/7)) f(θ)=(√(25(49+64))) cos (θ−α)+53 { ((f(θ)_(max) = 5(√(113)) +53)),((f(θ)_(min) = −5(√(113)) +53 )) :}
$$\mathrm{let}\:\begin{cases}{\mathrm{x}−\mathrm{3}=\mathrm{5cos}\:\theta\rightarrow\mathrm{x}=\mathrm{3}+\mathrm{5cos}\:\theta}\\{\mathrm{y}−\mathrm{4}=\mathrm{5sin}\:\theta\rightarrow\mathrm{y}=\mathrm{4}+\mathrm{5sin}\:\theta}\end{cases} \\ $$$$\mathrm{let}\:\mathrm{7}\left(\mathrm{3}+\mathrm{5cos}\:\theta\right)+\mathrm{8}\left(\mathrm{4}+\mathrm{5sin}\:\theta\right)=\mathrm{f}\left(\theta\right) \\ $$$$\mathrm{f}\left(\theta\right)=\:\mathrm{35cos}\:\theta+\mathrm{40sin}\:\theta+\mathrm{53} \\ $$$$\mathrm{f}\left(\theta\right)=\sqrt{\mathrm{35}^{\mathrm{2}} +\mathrm{40}^{\mathrm{2}} }\:\mathrm{cos}\:\:\left(\theta−\alpha\right)+\mathrm{53} \\ $$$$\mathrm{where}\:\alpha\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{8}}{\mathrm{7}}\right) \\ $$$$\mathrm{f}\left(\theta\right)=\sqrt{\mathrm{25}\left(\mathrm{49}+\mathrm{64}\right)}\:\mathrm{cos}\:\left(\theta−\alpha\right)+\mathrm{53} \\ $$$$\begin{cases}{\mathrm{f}\left(\theta\right)_{\mathrm{max}} \:=\:\mathrm{5}\sqrt{\mathrm{113}}\:+\mathrm{53}}\\{\mathrm{f}\left(\theta\right)_{\mathrm{min}} \:=\:−\mathrm{5}\sqrt{\mathrm{113}}\:+\mathrm{53}\:}\end{cases} \\ $$$$ \\ $$
Answered by mr W last updated on 06/May/21
an other way: let 7x+8y=k ⇒y=((k−7x)/8) (x−3)^2 +(((k−7x)/8)−4)^2 =25 113x^2 −(14k−64)x+k^2 −64k=0 such that x∈R exists, Δ=(14k−64)^2 −4×113×(k^2 −64k)≥0 ⇒k^2 −106k−16≤0 k_(1,2) =53±5(√(113)) ⇒53−5(√(113))≤k≤53+5(√(113)) i.e. k_(min) =53−5(√(113)) k_(max) =53+5(√(113))
$${an}\:{other}\:{way}: \\ $$$${let}\:\mathrm{7}{x}+\mathrm{8}{y}={k} \\ $$$$\Rightarrow{y}=\frac{{k}−\mathrm{7}{x}}{\mathrm{8}} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left(\frac{{k}−\mathrm{7}{x}}{\mathrm{8}}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$\mathrm{113}{x}^{\mathrm{2}} −\left(\mathrm{14}{k}−\mathrm{64}\right){x}+{k}^{\mathrm{2}} −\mathrm{64}{k}=\mathrm{0} \\ $$$${such}\:{that}\:{x}\in{R}\:{exists}, \\ $$$$\Delta=\left(\mathrm{14}{k}−\mathrm{64}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{113}×\left({k}^{\mathrm{2}} −\mathrm{64}{k}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{k}^{\mathrm{2}} −\mathrm{106}{k}−\mathrm{16}\leqslant\mathrm{0} \\ $$$${k}_{\mathrm{1},\mathrm{2}} =\mathrm{53}\pm\mathrm{5}\sqrt{\mathrm{113}} \\ $$$$\Rightarrow\mathrm{53}−\mathrm{5}\sqrt{\mathrm{113}}\leqslant{k}\leqslant\mathrm{53}+\mathrm{5}\sqrt{\mathrm{113}} \\ $$$${i}.{e}. \\ $$$${k}_{{min}} =\mathrm{53}−\mathrm{5}\sqrt{\mathrm{113}} \\ $$$${k}_{{max}} =\mathrm{53}+\mathrm{5}\sqrt{\mathrm{113}} \\ $$

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