What is the answer for this equation: 3n-8=n
2 answers:
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Answer: (A)
(B) Length varies between 1 and 150
(C) Largest area is 22500ft²
Step-by-step explanation: Suppose length is l and width is w.
The rectangular garden has perimeter of 600ft, which is mathematically represented as
Area of a rectangle is calculated as
Now, we have a system of equations:
Isolate w, so we have l:
w = 300 - l
Substitute in the area equation:
A = l(300 - l)
A = 300l - l²
(A) <u>Function of area in terms of length is given by </u><u>A = 300l - l²</u>
(B) The practical domain for this function is values between 1 and 150.
(C) For the largest area, we need to determine the largest garden possible. For that, we take first derivative of the function:
A' = 300 - 2l
Find the values of l when A'=0:
300 - 2l = 0
2l = 300
l = 150
Replace l in the equation:
w = 300 - 150
w = 150
Now, calculate the largest area:
A = 150*150
A = 22500
<u>The largest area the fence can enclose is </u><u>22500ft².</u>
Your calculator's cubic regression function can tell you the equation is
... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)
The x-intercepts are -4, 0, +1.5.
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If you want to solve this "by hand", you can first of all recognize that since there is an x-intercept at 0, the cubic will only have three coefficients. That is, you can write the equation as
... f(x) = ax³ + bx² + cx
Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.
... -a +b -c = 15 . . . . . for x=-1
... a + b + c = -5 . . . . for x=1
... 8a +4b +2c = 12 . . for x=2
adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to
... a + c = -10
... 4a +c = -4
Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).
Then your equation is
... f(x) = x(2x² +5x -12)
Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.
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Since the input values are sequential, you can also develop the function from differences of the output values. Those are 15, 0, -5, 12. First differences are -15, -5, +17. Second differences are +10, +22. The third difference is 12. Using the first of these differences in appropriate places in the interpolating polynomial formula, we have
... f(x) = 15 + (x+1)(-15 + (x)/2·(10 + (x-1)/3·(12))) = 2x³ +5x² -12x . . . . as above
