You are creating a rectangular garden and want to surround it with brick edging.

Question

You are creating a rectangular garden and want to surround it with brick edging. You also are going to divide it into two halves with this edging so that your vegetables and herbs are on one side and flowers are on the other. See diagram below for the general shape. You only have 120 feet worth of brick edging, and want to enclose the most area possible. Determine the dimensions (length and width) that will maximize the total area of the garden, state that maximum area. Define variables that you will use in the problem, including units. (You may use the diagram above to demonstrate.) Write equations or functions relating these variables and using the given information. Develop a function of one variable that describes the area of the garden and use calculus to locate the critical value(s). Check to find the critical value that is the maximum's location. (Show calculus work of some type) What are the best dimensions for the garden? __________ by __________. What is the garden area obtained by using these dimensions? __________ For the function below, do the following 3 things. Calculate the x-values of all critical points of the 1st derivative Determine, showing calculus work, whether the function increases or decreases on each interval of the domain around and between these critical values. Identify each critical value as the location of a local min, a local max or neither. f(x) = -x3 + 3x2 + 24x 7

Explanation / Answer

1 . let x is half of length and y is width

thus

2. area = 2x*y = 2xy

given = 2 (2x+y) = 120

2x+y = 120/2 = 60

y = (60-2x)

3. area = 2xy =    2x*(60-2x)

take derivative

da/dx = 2x *-2 + (60-2x)*2 = 0 => -8x+120 = 0

-4x + 120 -4x = 0

120 = 8x

x= 120/8 = 15

y = 60-2x = 60 - 2*15 = 60-30 = 30

4. take double derivative

d^a/dx^2 = -8 < 0

so at x = 15 it has max value.

5. 2x by y = 2*15 by 30 = 30 by 30

garden area = 30 by 30


1. f(x) = -x^3 + 3x^2 +24x - 7

f'(x) = -3x^2 +6x + 24 = 0

-3 (x^2 -2x-8) = 0

x^2 -4x+2x-8 =0

x(x-4) +2(x-4) = 0

(x+2)(x-4) = 0

x+2 = 0 or x = -2
x-4 =0 or x= 4

critical values x = -2, 4

b)
between (infiny to -2) select x = -3

between (-2 to 4) select x = 0

between (4, infinity) x = 3

f'(-3) = -3*(-3+2)(-4-3) = -1*-7*-3 = -3*7 = -21 <0

between (infiny to -2)    function is decreasing

f'(0) = -3*(2)*-4 = -3* -8 = 24 > 0

between ( -2 to 4 )    function is increasing

f(3) = -3*(3+2)(3-4) =-3* 5*-1 = 15 > 0

between (4 to infiny to )    function is increasing

c) f''(x) = -6x +6

f''(-2) = -6*-2+6 = 18 > 0   function has min at x = -2

f''(4) = -6*4+6 = -18 < 0 function has max at x = 4

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