optimisation higher maths

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maxima OPTIMISATION minima Ex A rectangular area, has to be fenced off with a total length of 280m of fencing. a) Given that x is the length of the rectangle, show that the area is given by A (x) = 140x-x² x x 19 Area = xy Perimeter = 2x +2y Watch for factorising A'(x)=140-2x For st pts (et A'(x) = 0 140-2x = 0 140=2x x=70 y=140-x Hink for a MAX b)Hence find the greatest area that can be enclosed and the dimensions of the rectangle. A(x) = 140x-x² x60 70 80 A') + O Slope V=lxbxh 108=x*x*y 108=x²y y=10€ S(x)=x²+432 S(x)=x²+432x" S'(x)=2x-432x=² S'(x) = 2x - 432 For st pts let s'(x) = 0 * 2x - 432=0 Perimeter = 280m 280=2x+2y 2y=-2x+280 Ex2 Open rectangular bins with square bases and a volume of 108m² are to be made using a minimum of metal. If x metres is the length of a side of the base and height of each bing show that the total surface area of each bin is given by S(x) = x² + 43²282 6) Calculate the dimensions of then bin so the metal used is a minimum. metres is the y Total SA 5(x)=x² +- 2x = 432 2x³=432 x³ =216 x = = 3√216 x=6 Base = xxx = x² Sides = 4* x*y * S = 4x+10 x 5¹(x) Slope S= 432 x 1 610 - 0 :.Max at x = 70 Area =x (140-x) A(x) = 140x-x² +y=108 • min at x = 6 Use in area Find dimensions At x=70 A(x) = 140x-x² A (70) - 140x70-(70) ² = 9800 - 4900 = 4900 • The greatest...

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