Optimization Question

Question

A "cross" is inscribed in a circle of radius 9 (see diagram below). The cross is symmetric, so each outer edge (the ones in purple) has the same length, say 2x. We seek to find x so that the area of the cross is maximized. You may move the slider to see the effect on the cross when x is increased or decreased.

(a) Write the function A(x) that gives the area of the cross as a function of x. Let 2x be the length of an outer edge of the cross (the segments in purple).

(b) What is the open interval on which A(x) is defined? We'll insist that the interval is open so that we actually have a cross.

(c) A′(x)= 

(d) What is the critical number for A(x) that lies in the interval identified in part (b)? Write an exact value, i.e. no decimal approximation.

Answer 1

max cross area = 162 when x=sqr(9^2/2)= sqr40.5

max area =4sqr(81-x^2)^2= 4(40.5)=162

when the cross degenerates into totally overlapping squares

sqr162 by sqr162= 9sqr2 by 9sqr2=

about 9(1.414) by 9(1.414)= about 12.73 by 12.73

if x= 0, Area = 0 when the cross degernerates into 2 perpendicular lines of zero width

domain of the Area = (0,162)

domain of x = (0,sqr40.5)= about (6.364)

circle is x^2+y^2=9^2=81

A(x)= 4xsqr(81-x^2)

A'(x)= 4x(-2x)/sqr(81-x^2)+ 4sqr(81-x^2)=0

-2x^2 +81-x^2=0

3x^2=81

x^2=81/3=27

x=9sqr3

a slight mistake somewhere, if max x=9sqr2/2=about 6.364