數學問題 40分

.A manufacturer finds that he can sell x refrigerators permonth at $y each where 0≦x≦2000 and y=-0.038x+800. Thecost $C of producing x refrigerators per month is given by C=580+620x-0.003x平方. Find the value of x forwhich the profit is the greatest.The Revenue = selling price per unit multiplied # of unitssoldThe Revenue = (- 0.0338x + 800)xThe cost C =580 +620x -0.003x^2Profit = Total Revenue – Total CostP = (- 0.0338x + 800)x – (580 +620x -0.003x^2)       - 0.0338x^2 +800x –580- 620x +0.003x^2       - 0.0308x^2 +180x –580Differentiate dP/dx dP/dx = -0.00616x +180 = 0 (Put dP/dx = 0)0.00616x = 180x = 29220x = 29220 for which the profit is the greatest.Comment: x = 29220 is outside the range 0≦x≦2000Please CHECK the figures and make necessary correction ifthe input figure is changed.2.A rectangle is to be inscribed in the circle x平方+y平方=36. Find the largest capacity.The radius of the circle Square root of 36 = 6The rectangle inscribed is (2x) in length and (2y) in widthwhere x^2 + y^2 = r^2 where r is the radius of circle, x is from the centre ofrectangle to the width and y the centre of rectangle to the lengthx^2 + y^2 = r^2,  y^2= r^2 - x^2 y  = square root (r^2- x^2)Area of rectangle = (2x)(2y)A(x) = 4x square root (r^2 - x^2)Differentiate dA/dx = 4(sq. root(r^2 - x^2) x^2 –x^2/( sq.root(r^2 - x^2))4(r^2 – 2x^2)_______________     = 0 (dA/dx = 0)sq. root(r^2 - x^2)The critical number is x = r/sqroot 2. Clearly gives a maximum.The dimension 2x = r (sq. rootof 2) and 2y = r (sq. root of 2)The rectangle is a square inthis caseArea = (2x)(2y)r (sq. root of 2) r (sq. root of2)A = 2r^2 = 2(6)^2 (radius of circle = 6) A = 72 sq. unit

可用 x^2 表示 x 的平方

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