How do you determine the values of the coefficients a and b that will give a minimum value of the area, #S#, of the region enclosed by the curve #y=-x^2+ax+b# (which passes through point (1,2)) and the curve #y=(1/2)x^2#?

1 Answer
Cesareo R.
Dec 11, 2016

#S=2#

Explanation:

Calling

#f_1(x)=-x^2+ax+b#
#f_2(x)=1/2x^2#

the intersection points between #f_1# and #f_2# are obtained solving

#{(f_1(1)=2->-1+a+b=2),(f_1(x)=y->-x^2+ax+b=y),(f_2(x)=y->1/x^2=y):}#

Solving for #x,y,a# we obtain

#((x = 1/3 (3 - b - sqrt[9 + b^2]), y = 1/9 (9 - 3 sqrt[9 + b^2] + b (-3 + b + sqrt[9 + b^2]))), (x = 1/3 (3 - b + sqrt[9 + b^2]), y = 1/9 (b^2 + 3 (3 + sqrt[9 + b^2]) - b (3 + sqrt[9 + b^2]))))#

with #a=3-b#

so the intersection points are at

#x_i=1/3 (3 - b - sqrt[9 + b^2])#
#x_s=1/3 (3 - b + sqrt[9 + b^2])#

The enclosed area is

#S(b)=int_(x=x_i)^(x=x_s)(f_1(x)-f_2(x))dx = 1/2 (x_i - x_s) (-2 b + x_i^2 + x_i x_s + x_s^2 - (3-b)(x_i + x_s))#

Substituting now #x_i# and #x_s# we obtain

#S(b)=2/27 (9 + b^2)^(3/2)# with a minimum at #b=0#

So finally #a=3,b=0# and #S=2#

Attached a plot showing #f_1(x)# blue and #f_2(x)# red.

enter image source here

Related questions
See all questions in Integration: the Area Problem
Impact of this question
2023 views around the world
You can reuse this answer
Creative Commons License