Question

Consider the two functions:

f(x,a) =70- 5xa & g(x) = 30 +4x

a) find the x value of the point where the two equations intersect (in terms of the variable a)

b) Find the value of the functions at the point where the two equations intersect (in terms of the variable a).

c) Take the partial derivate of f with respect to x, \partial f / \partial x, and with respect to a, \partial f / \partial a

d) What are the values of these derivatives , when x= 3 and a=2, which can be written as \partialf /\partialx (3,2) and \partial f /\partiala (3,2)

e) Next caculate these two numbers:

\upsilon1 = (\partialf /\partialx ( 3,2)) 3 / f (3,2)

\upsilon2 = (\partialf /\partiala ( 3,2)) 2 / f (3,2)

f) Finally write out these equations in term of a and x and simplify,

\upsilon1 = (\partialf /\partialx ( x,a)) x / f (x,a)

\upsilon2 = (\partialf /\partialx ( x,a)) a / f (x,a)

Answer 1

a) The x value of the point where the two equations intersect in terms of a is
`x=(40)/(4+5a)`

b) The value of the functions at the point where they intersect is
`(10 (28 + 15 a))/(4 + 5 a)`

c) The partial derivative of f with respect to
`x` is
`(\partial f)/(\partial x) = -5a` and the partial derivative of f with respect to
`a` is
`(\partial f)/(\partial x) = -5x`

d) The value of
`(\partial f)/(\partial x)(3,2) = -10` and
`(\partial f)/(\partial a)(3,2) = -15`

e)
`\upsilon_1=-(3)/(4) = -0.75` and
`\upsilon_2=-(3)/(4) = -0.75`

f) equation
`\upsilon_1 = (-5a\cdot x)/(70-5ax)=(ax)/(ax-14)` and
`\upsilon_2 = (-5a\cdot a)/(70-5ax)=(a^2)/(ax-14)`

Explanation:

a) In order to find the
`x` we just need to equal the equations and solve for
`x`:

`f(x,a)=g(x)\\70-5xa = 30+4x\\70-30 = 4x+5xa\\40 = x(4+5a)\\\boxed {x = (40)/(4+5a)}`

b) Since we need to find the value of the function in the intersection point we just need to substitute the result from a) in one of the functions. As a sanity check , I will do it in both and the value (in terms of
`a`) must be the same.

`f(x,a)=70-5ax\\f((40)/(4+5a), a) = 70-5\cdot a \cdot  (40)/(4+5a)\\f((40)/(4+5a), a) = 70 - (200a)/(4+5a)\\f((40)/(4+5a), a) = (70(4+5a) -200a)/(4+5a)\\f((40)/(4+5a), a) =(280+350a-200a)/(4+5a)\\\boxed{ f((40)/(4+5a), a) =(10(28+15a))/(4+5a)}`

and for
`g(x)`:

`g(x)=30+4x\\g((40)/(4+5a))=30+4\cdot (40)/(4+5a)\\g((40)/(4+5a))=(30(4+5a)+80)/(4+5a)\\g((40)/(4+5a))=(120+150a+80)/(4+5a)\\\boxed {g((40)/(4+5a))=(10(28+15a))/(4+5a)}`

c)
`(\partial f)/(\partial x) = (70-5xa)^(')=70^(') - (\partial (5xa))/(\partial x)=0-5a\\(\partial f)/(\partial x) =-5a`

`(\partial f)/(\partial a) = (70-5xa)^(')=70^(') - (\partial (5xa))/(\partial a)=0-5x\\(\partial f)/(\partial a) =-5x`

d) Then evaluating:

`(\partial f)/(\partial x) =-5a\\(\partial f)/(\partial x) =-5\cdot 2=-10`

`(\partial f)/(\partial a) =-5x\\(\partial f)/(\partial a) =-5\cdot 3=-15`

e) Substituting the corresponding values:

`\upsilon_1 = (\partial f(3,2))/(\partial x)\cdot (3)/(f(3,2)) \\\upsilon_1 = -10 \cdot (3)/(40)  = -(3)/(4) = -0.75`

`\upsilon_2 = (\partial f(3,2))/(\partial a)\cdot (3)/(f(3,2)) \\\upsilon_2 = -15 \cdot (2)/(40)  = -(3)/(4) = -0.75`

f) Writing the equations:

`\upsilon_1=(\partial f (x,a))/(\partial x)\cdot (x)/(f(x,a))\\\upsilon_1=-5a\cdot (x)/(70-5xa)\\\upsilon_1=(-5ax)/(70-5ax)=(-5ax)/(-5(ax-14))\\\boxed{\upsilon_1=(ax)/(ax-14) }`

`\upsilon_2=(\partial f (x,a))/(\partial x)\cdot (a)/(f(x,a))\\\upsilon_2=-5a\cdot (a)/(70-5xa)\\\upsilon_2=(-5a^2)/(70-5ax)=(-5a^2)/(-5(ax-14))\\\boxed{\upsilon_2=(a^2)/(ax-14) }`