# Assignment # 6 - SOLUTIONS 
#  -- One Variable Optimization --
# 
# Question 1:
> f:=2*x^3+3*x^2-36*x+1;
> df:=diff(f,x);
> solve(df=0,x);
# The possible max/min are at x = -3 and x = 2.
> plot([f,df],x=-5..5,color=[red,blue]);
# We can see (from the graph and from "solve") that the local max of
# f(x) is at x = -3 and the local min of f(x) is at x = 2.  We can also
# determine that f(x) is increasing when 
> `<`(x, `+`(`-`(3)));#  and 
> `<`(x, 2);# , and f(x) is decreasing when -3 < x < 2.
# 
# Question 2:
> f:=3*x^5-10*x^4+7*x;
> df:=diff(f,x);
> df2:=diff(f,x,x);
> solve(df2=0,x);
# The possible inflection points are at x = 0 and x = 2.
> plot([f,df2],x=-1..3,color=[red,blue]);
# We can see from the above graph that f(x) is concave up when x > 2 and
# concave down when x < 2.  Note: x = 0 is NOT an inflection point.
# 
# Question 3:
# a) Let us determine the derivatives of y:
> y:=x^3-3*x^2-9*x+10;
> yprime:=diff(y,x);
> y2prime:=diff(y,x,x);
> solve(yprime=0,x);
# so, the possible max/min are at x = -1 and x = 3.
> plot([y,yprime],x=-2..4, color=[red,blue]);
# b) We can tell from the above graph that y is increasing when x < -1
# and x > 3 and y is decreasing when -1 < x < 3.
# c) We can also tell, from the above graph that y has a relative
# maximum at x = -1 and a relative minimum at x = 3.
# d) To find where y is concave up and concave down, let us plot y and
# y'':
> solve(y2prime=0,x);
> plot([y,y2prime],x=-2..4,color=[red,blue]);
# We can determine, from the above graph that y is concave up when x > 1
# and concave down when x < 1.
# e) There is one inflection point (as we can see from the graph) at x =
# 1.
# f) See the above graphs of y.
# 
# Question 4:
> f:=(x^2-3*x+2)^2;
> fprime:=diff(f,x);
> solve(fprime=0,x);
> plot([f,fprime],x=0.5..2.5,color=[red,blue]);
# As we can see from the above plots, the local minima are at x = 1 and
# x = 2 and the local maxima is at x = 1.5.  We can also determine that
# f is increasing when 1 < x < 1.5 and x > 2.  Also, f is decreasing
# when - 1 < x and 1.5 < x < 2.
# 
# Question 5:
> y:=x^3+4*x^2-3*x+4;
> yprime:=diff(y,x);
> y2prime:=diff(y,x,x);
> solve(yprime=0,x);
> subs(x=1/3,y2prime);
# Since y''(x) is positive at x = 1/3, then there is a local minimum
# there.
> subs(x=-3,y2prime);
# Since y''(x) is negative at x = - 3, then there is a local maximum
# there.
> plot(y,x=-4..2);
# 
# 
# Question 6:
> C:=2*x^3-15*x^2-84*x+3100;
> Cprime:=diff(C,x);
> C2prime:=diff(C,x,x);
> solve(Cprime=0,x);
> subs(x=-2, C2prime);
# Since C''(-2) < 0, then x = -2 is a local maximum. BUT! x = -2 is NOT
# possible, since you cannot have a negative quantity.
> subs(x=7, C2prime);
# Since C''(7) > 0, then x = 7 is a local minimum.
# 
# 
# Question 7:
> C:=2*x^3-39*x^2+180*x+21200;
> Cprime:=diff(C,x);
> C2prime:=diff(C,x,x);
> solve(Cprime=0,x);
> subs(x=3, C2prime);
# Since C''(3) < 0, then x = 3 is a local maximum.
> subs(x=10, C2prime);
# Since C''(10) > 0, then x = 10 is a local minimum.
# 
# 
# Question 8:
> p:=200-0.98*q;
> c:=0.02*q^2+2*q+8000;
> P:=p*q-c;
> P:=simplify(P);
> Pprime:=diff(P,q);
> P2prime:=diff(P,q,q);
> solve(Pprime=0,q);
# So, the profit is maximized when q (quantity) = 99 units.
# 
# Question 9:
> p:=100-3*q;
> c:=(4+100/q)*q;
> c:=simplify(c);
> P:=p*q-c;
> P:=simplify(P);
> Pprime:=diff(P,q);
> P2prime:=diff(P,q,q);
> solve(Pprime=0,q);
# So, the profit is maximized when quantity (q)  = 16 units.
# 
# Question 10:
> p:=500/sqrt(q);
> c:=5*q+2000;
> P:=p*q-c;
> P:=simplify(P);
> Pprime:=diff(P,q);
> P2prime:=diff(P,q,q);
> solve(Pprime=0,q);
# So, the profit is maximized when q = 2500 units.
#