Question'answers seminar 1 Answers
9/24/2010
Bus 14B – Question and Answer Seminar #1
(Fall ‘10)
A1)
Maximizing Word Problem – Solve by Algebra
Find the dimensions of a rectangular field of
maximum
area
that can be made from 480 m of fencing material.
(See 14.2)
Given:
Perimeter so 2x + 2y = 480
Goal
:
Maximize area of field
Area = x * y
A(x) = x *
y
A(x) = x
( 240 – x)
A(x) = x
2
+ 240x
This is a parabola opening
downward.
Highest or max
point is the vertex point.
X value at the vertex point
Is the axis of symmetry.
X = B/(2A)
X = 240/2 = 120 m.
To find y, substitute into perimeter equation.
Y = 120m.
Dimensions are 120 x 120.
A2)
To Solve for Maximum or Minimum by Calculus
a)
Take the Derivative of the function.
b)
Set derivative equal to zero.
c)
Solve for “x”.
Derivative refers to the slope of curve,
instantaneous rate of change, slope of tangent line at
point on the curve.
Symbols for the derivative:
y’, f’(x),
x
y
∆
∆
The formula below is used to find the derivative.
But in Ch. 12,
you are given many rules.
The first one, power rule, is shown in
the next section.
It is much easier to use the rules, than do use
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 Spring '08
 DONTKNOW
 Calculus, Derivative, Power Rule, Limit, lim

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