When what I say clearly does not apply to the problem, you should tell me. Here I assigned Problem 60 in Section 7.5 when I meant Section 7.7. I’ve included the answer for the problem in Section 7.5, even though it’s a pointless problem.
x  80  85  90  95  100 
y  4000  2050  0  2150  4400 
So the zero is at x = 90. Plotting these segments:
See PDF version.
For using points, find convenient numbers in the problem and try them. Here, to numbers that pop out are 30 and 150. Evaluating at each gives (30,14400) and (150,50400), so we know the zero must be somewhere between them. Halfway is (30 + 150)∕2 = 90, so trying 90 finds the solution.
{x  1980\over
1990  1980} = {y  4.5\over
5.2  4.5}.

Plugging in x = 1985 and solving for y gives y = \mathbf{4.85} million. Repeating around 1995,
{x  1990\over
2000  1990} = {y  5.2\over
5.8  5.2}.

Using x = 1995 gives y = \mathbf{5.5} million.
{x  150\over
1400  150} = {y  5000\over
24000  5000}.

Solving for y gives the slopeintercept form,
y = 15.2x + 2720.

To verify this, check that x = 150 gives y = 5000, and x = 1400 gives y = 24000.
{x  5\over
7  5} = {y  24075\over
26628  24075}.

Solving for y,
This is best solved by substitution. Written in terms of C, the last equation becomes (C  10) + C = 490, or \mathbf{C = 250}. Now writing the first in terms of A after substituting C, 250 = A + 2A + 10 or \mathbf{A = 80}. Now \mathbf{B = 160}. Substituting these into the equations above verifies this solution.
Product  I (g/serving)  II (g/serving)  cost ($/serving) 
A  3  2  0.25 
B  2  4  0.40 
So the function to minimize is the cost 0.25A + 0.40B. The constraints are that there be at least 15 g of I, or 3A + 2B ≥ 15, and 15 g of II, or 2A + 4B ≥ 15. There also are trivial constraints A ≥ 0 and B ≥ 0. The problem to solve is to
The first two lines intersect at (3.75,1.875). The A intercepts are 5 and 7.5, and only (0,7.5) is feasible. The B intercepts are 7.5 and 3.75, and only (7.5,0) is feasible. So the points to check and the function values are
Point  Cost 
(3.75,1.875)  $1.6875 
(0,7.5)  $3 
(7.5,0)  $1.875 
So the cheapest combination is 3.75 of A and 1.875 of B at $1.6875.
Kind  Variable  Oven time (hr)  Decorating time (hr)  Profit ($) 
Cookie  C  1.5  {2\over 3}  20 
Cake  A  2  3  30 
So the problem is to
The interesting points and their values:
Point  Profit ($) 
(0,0)  0 
(6,3)  210 
(0, {13\over 3} )  130 
(10,0)  200 
So the best combination is 6 cookie batches and 3 cake batches for a profit of $210.