ChIIISVNonLinearSystems052 - 3.5: NonLinear Systems Dr....

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
3.5: NonLinear Systems Dr. Raja Mohammad Latif
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Math 131-052. Dr. Raja Latif 2 1 3.5 Nonlinear Systems ===Lecture Sec 3.5 Begins now=== The method of substitution is often useful when we have a system of equations in which one equation is linear and the other is not. 150AL5Example. Solve the following system of equations 2x y = 3 x 2 + y 2 = 5 : Solution. In this system, one of the two equation is not linear. The method of solution still consists of elimination of one of the variables, x or y, from the two equations. y = 2x 3 : We substitute this value of y into the second equation and simplify. x 2 +( 2x 3 ) 2 = 5 x 2 + 4x 2 12x + 9 = 5 5x 2 12x + 4 = 0 This factors to give ( x 2 )( 5x 2 ) = 0 : Therefore either x 2 = 0or5x 2 = 0 x = 2orx = 2 = 5 : Dept Maths. KFUPM
Background image of page 2
Math 131-052. Dr. Raja Latif 3 Next, we substitute these values into the equation we used earlier to substitute for y, namely y = 2x 3 : Thus x = 2 = ) y = 2 ( 2 ) 3 = 1 ; x = 2 = 3 = ) y = 2 ( 2 = 3 ) 3 = 5 = 3 Hence there are two solutions, x = 2 ; y = 1 and x = 2 = 3 ; y = 5 = 3 : ===================== 41HB26. When electric blenders are sold for p dollars a piece, manufacturers will supply p 2 10 blenders to local retailers, while the local demand will be 60 - p blenders. At what market price will the manufacturers, supply of blenders be equal to the consumers’ demand? Solution. We set the supply and demand equations equal to each other and solve. Supply = Demand p 2 10 = 60 p = ) 10 ( p 2 10 ) = 10 ( 60 p ) Dept Maths. KFUPM
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Math 131-052. Dr. Raja Latif 4 = ) p 2 = 600 10p = ) p 2 + 10p 600 = 0 = ) ( P + 30) ( p 20) = 0 = ) p = 20 ; 30 Reject 30 since p ± 0 : Thus, the equilibrium price is $ 20 : ========================== 162AL7.(Nonlinear Break-even Analysis). The cost of producing x items per day is given in dollars by y c = 80 + 4x + 0 : 1x 2 : If each item can be sold for $ 10 , determine the break-even point. Solution. The cost is given by y c = 80 + 4x + 0 : 1x 2 : If each item sells for $ 10 , the revenue (in dollars) obtained by the sale of x items is y R = 10x. At the break-even point, we have y c = y R ; that is, 80 + 4x + 0 : 1x 2 = 10x : = ) 0 : 1x 2 6x + 80 = 0 = ) x 2 60x + 800 = 0 Dept Maths. KFUPM
Background image of page 4
Math 131-052. Dr. Raja Latif 5 = ) ( x 20 )( x 40 ) = 0 = ) x = 20 or x = 40 : The break-even point is 20 or 40 items. =============== 166AL39. (Market Equilibrium and Revenue). The
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/17/2009 for the course MATH MATH131 taught by Professor Dr.rajalatif during the Spring '09 term at King Fahd University of Petroleum & Minerals.

Page1 / 14

ChIIISVNonLinearSystems052 - 3.5: NonLinear Systems Dr....

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online