*20 - 3 Extreme
value
problems*
**This section deals
with
relating maximum and minimum problems to your favorite type of
problem.
Many business problems relate directly to maximizing profits while
minimizing
expenses. This section deals with simple examples of these types
of problems.**
*Sample problems*
*1) What is
the
area of the largest rectangular garden that a farmer can enclose with
500
feet of fencing if one side of the garden will be the side of a barn?*

*A(x) = L*^{ .}
W
*A(x) = x(500 - 2x)
=
500x - 2x*^{2}
*To find the maximum
value,
find the derivative of A(x)*
*A'(x) = 500 - 4x*
*Set it equal to
zero
and solve!*
*500 - 4x = 0*
*-4x = -500*
*x = 125*
*To have a maximum
area
of the garden, the width must be 125' and the length must be 250'.*
*Therefore, the area
is
maxed at (125)(250) = *__31250 square feet__!
*2) A
manufacturer
has 100 tons of metal that he can sell now with a profit of $5 a
ton.
For each week that he delays shipment, he can produce another 10 tons
of
metal. However, for each week he waits, the profit drops 25 cents
a ton. If he can sell the metal at any time, when is the best
time
to sell so that his profit is maximized?*
*Let x = the
number of
weeks to wait!*
Ship |
Amount of metal |
Profit per ton |
Total profit |

now |
100 |
5 |
500 |

in x weeks |
(100 + 10x) |
(5 - .25x) |
500 +25x -2.5x^{2} |

*P(x) = 500 + 25x
- 2.5x*^{2}
*Find the
derivative!*
*P'(x) = 25 - 5x*
*Set P'(x) = 0 and
solve*
*25 - 5x = 0*
*-5x = -25*
*x = 5*
*Best time to sell
is
in 5 weeks. He will have 150 tons and the price will be $3.75 for
a maximum profit of *__$562.50__.
*3) Given a
20-unit
square of sheet metal, find the dimensions of an open box ( no top) of
greatest volume that can be made by cutting congruent squares from the
corners.*

*Use the diagram
above
for the metal sheet. The corners are being folded up to form the
box. V(x) = L*^{ .} W ^{.} H
*V(x) = x(20 - 2x)*^{2}
= 400x - 80x^{2}
+ 4x^{3}
*Find the derivative
to
find the maximum point!*
*V'(x) = 400 - 160x
+
12x*^{2}
*= 4(100 - 40x + 3x*^{2})
*= 4(10 - 3x)(10 - x)*
*We have two
critical
points: x = 10 and x = 10/3*
*What happens when x
=
10! The volume will be zero because you are cutting the metal in
half. You won't be able to form a box and the volume is
zero.
This is a minimum value. Thus the maximum value happens at*
__x = 10/3__
*The volume =
10/3(20
- 20/3)*^{2} = __16,000/27
cubic units!__
* *
*Really hate word
problems
don't you!!*