LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbW9HRiQ2LVEhRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0Y0LyUpc3RyZXRjaHlHRjQvJSpzeW1tZXRyaWNHRjQvJShsYXJnZW9wR0Y0LyUubW92YWJsZWxpbWl0c0dGNC8lJ2FjY2VudEdGNC8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRkMvJStleGVjdXRhYmxlR0Y0Ri8= # Katie Migatulski restart; #Problem 1: a. The length of one side of the square in terms of x is ....(10-x)/4 #The square half of the wire is denoted as 10-x # b. The radius of the circle is...(x)/(2Pi) #Endpoints=[0,10] A:= x->(x^2/(4*Pi))+((x^2-20*x+100)/16); #Function of Total Area Zio2I0kieEc2IkYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCoqJjkkIiIjSSNQaUclKnByb3RlY3RlZEchIiIjIiIiIiIlKiRGK0YsI0YxIiM7RisjISImRjIjIiNERjJGMUYlRiVGJQ== diff(A(x),x); #Taking the derivative of the Total Area function LCgqJkkieEc2IiIiIkkjUGlHJSpwcm90ZWN0ZWRHISIiI0YmIiIjRiQjRiYiIikjISImIiIlRiY= a:=x->((1/2)*x/Pi+(1/8)*x-5/4); Zio2I0kieEc2IkYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCgqJjkkIiIiSSNQaUclKnByb3RlY3RlZEchIiIjRiwiIiNGKyNGLCIiKSMhIiYiIiVGLEYlRiVGJQ== solve((1/2)*x/Pi+(1/8)*x-5/4=0,x); #Set the Derivative = 0 and find Critical Point LCQqJkkjUGlHJSpwcm90ZWN0ZWRHIiIiLCZGJEYmIiIlRiYhIiIiIzU= #Critical Point= 10*Pi/(Pi+4) A(0); #Checking for global min. / max.@ endpoint A(10); #Checking for global min. /max. @ endpoint IyIjRCIiJQ== LCQqJEkjUGlHJSpwcm90ZWN0ZWRHISIiIiNE evalf(25/4); #Evaluating A(0) as a decimal JCIrKysrXWkhIio= evalf(25/Pi); #Evaluating A(10) as a decimal JCIrX3J1ZHohIio= evalf(10*Pi/(Pi+4)); #Evaluating Critical Point as a decimal JCIrbCUzISpSJSEiKg== #The largest total area is 7.96 ft.^2 when you cut the wire at x= 10 ft. so if you cut a very small portion of the wire at the end close to x=10 ft., then almost the entire length of the wire is a part of the circle. #Problem 2: Suppose that a weight is to be held 10 feet below a horizontal line AB by a wire in the shape of a Y. If the points A and B are 8 ft apart, what is the shortest total length of the wire that can be used? #Used the pythagorean theorem to get x in terms of y so that I only use one variable. L := (y-> 2*sqrt(16+y^2)+10-y); #Created a function for the length of the wire Zio2I0kieUc2IkYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCgtSSVzcXJ0R0YlNiMsJiIjOyIiIiokOSQiIiNGL0YyIiM1Ri9GMSEiIkYlRiVGJQ== diff(L(y),y); #Taking the derivative of the length function LCYqJiwmKiRJInlHNiIiIiMiIiIiIztGKSMhIiJGKEYmRilGKEYsRik= l := (2*y/sqrt(y^2+16)-1); #The Derivative Function LCYqJiwmKiRJInlHNiIiIiMiIiIiIztGKSMhIiJGKEYmRilGKEYsRik= solve(l(y)=0,y); #Finding the critical point setting the derivative function = 0 Zio2I0kieUc2IkYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCQqJCIiJCMiIiIiIiMjIiIlRitGJUYlRiU= evalf((4/3)*sqrt(3)); #Evaluating the critical point JCIreDVTNEIhIio= evalf(L((4/3)*sqrt(3))); #Finding the length at the critical point JCIrQi4jR3AiISIp #Endpoint (0,10) limit(L(y),y=0); #Finding the limit as x approaches 0 IiM9 evalf(limit(L(y),y=10)); #Finding the limit as x approaches 10 JCIrQmYxYUAhIik= #The shortest length of the wire that can be used is 16.9 ft. LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbW9HRiQ2LVEhRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0Y0LyUpc3RyZXRjaHlHRjQvJSpzeW1tZXRyaWNHRjQvJShsYXJnZW9wR0Y0LyUubW92YWJsZWxpbWl0c0dGNC8lJ2FjY2VudEdGNC8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRkM=