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Question Number 68349 by peter frank last updated on 09/Sep/19

Answered by mr W last updated on 09/Sep/19

w_(new) =1.03w   (3% more=factor 1.03)  d_(new) =0.975d   (2.5% less=factor 0.975)  t_(new) =1.04t  (4% more=factor 1.04)  y=((Kwd^4 )/t^3 )  y_(new) =((Kw_(new) d_(new) ^4 )/t_(new) ^3 )  =((K1.03w0.975^4 d^4 )/(1.04^3 t^3 ))=0.827((Kwd^4 )/t^3 )  =0.827y=82.7% of y  100%−82.7%=17.3%  ⇒y is decreased by 17.3%

$${w}_{{new}} =\mathrm{1}.\mathrm{03}{w}\:\:\:\left(\mathrm{3\%}\:{more}={factor}\:\mathrm{1}.\mathrm{03}\right) \\ $$$${d}_{{new}} =\mathrm{0}.\mathrm{975}{d}\:\:\:\left(\mathrm{2}.\mathrm{5\%}\:{less}={factor}\:\mathrm{0}.\mathrm{975}\right) \\ $$$${t}_{{new}} =\mathrm{1}.\mathrm{04}{t}\:\:\left(\mathrm{4\%}\:{more}={factor}\:\mathrm{1}.\mathrm{04}\right) \\ $$$${y}=\frac{{Kwd}^{\mathrm{4}} }{{t}^{\mathrm{3}} } \\ $$$${y}_{{new}} =\frac{{Kw}_{{new}} {d}_{{new}} ^{\mathrm{4}} }{{t}_{{new}} ^{\mathrm{3}} } \\ $$$$=\frac{{K}\mathrm{1}.\mathrm{03}{w}\mathrm{0}.\mathrm{975}^{\mathrm{4}} {d}^{\mathrm{4}} }{\mathrm{1}.\mathrm{04}^{\mathrm{3}} {t}^{\mathrm{3}} }=\mathrm{0}.\mathrm{827}\frac{{Kwd}^{\mathrm{4}} }{{t}^{\mathrm{3}} } \\ $$$$=\mathrm{0}.\mathrm{827}{y}=\mathrm{82}.\mathrm{7\%}\:{of}\:{y} \\ $$$$\mathrm{100\%}−\mathrm{82}.\mathrm{7\%}=\mathrm{17}.\mathrm{3\%} \\ $$$$\Rightarrow{y}\:{is}\:{decreased}\:{by}\:\mathrm{17}.\mathrm{3\%} \\ $$

Commented by peter frank last updated on 09/Sep/19

please sir elaborate

$${please}\:{sir}\:{elaborate} \\ $$

Commented by peter frank last updated on 21/Sep/19

thank you

$${thank}\:{you} \\ $$

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