```Question 76379
You are given that:
.
{{{Y = A/X}}}
.
with the restrictions that
.
{{{2<=X<=6}}} and {{{(4/3)<=y<=B}}}
.
The problem is to find the value of both A and B
.
Go to the equation {{{Y = A/X}}} and let's solve for the range of Y based on the domain
of values that X can take.  From the restrictions on X we know that the smallest value
that X can take is 2 and the largest value that X can take is 6.  We will use these
two values to find the smallest and largest values that Y can be based on the values
of X.
.
If X is the smallest value it can be (2) then Y is:
.
{{{Y = A/2}}}
.
If X is the largest value it can be (6) then Y is:
.
{{{Y = A/6}}}
.
Now notice that {{{A/6}}} is smaller than {{{A/2}}} because the larger the denominator
the smaller the value of the fraction.  Therefore, the larger that X gets, the smaller that
Y will be.
.
So we can say that the smallest Y can be is {{{A/6}}} and the largest Y can be is {{{A/2}}}.
.
This can be translated to {{{A/6 <= Y <= A/2}}}.
.
Compare this to what limits on Y the problem gave ... {{{4/3<=Y<=B}}}
.
Now compare the left half of these two inequalities for Y and you can see that:
.
{{{A/6 = 4/3}}}
.
Solve this proportion by cross-multiplying the numerator of the left side by the denominator
of the right side. Then multiply the denominator of the left side times the numerator
of the right side. Then set these two products equal.  If you do this to this problem,
the two products are 3A and 24.  Set them equal and you get 3A = 24. Solve for A by dividing
both sides by 3 and you get A = 8.  That's one of the terms you were to find.
.
Now compare the right sides of the two inequalities we have for Y.  For them to be equivalent
you have to have {{{A/2 = B}}}.  But you now know that A = 8.  Substitute 8 for A and
you get:
.
{{{8/2 = B}}} and this simplifies to {{{B = 4}}}.
.
So the two answers you need are A = 8 and B = 4.
.
Hope that you can follow the above process and that it is understandable. The cross-multiplying
method of solving proportions is a method that can be used for all proportions, not just
for the one in this problem.```