I SWEAR this is my last one -- Algebra I problem?

Question

Carson has 140 m. of fencing to enclose a field. He wants to use sides that are only whole meter lenghts. What dimensions should he use to enclose the greatest possible area?

and

The sum of three consecutive integers is 378. What are the intergers?


yeahh, I'm doing my winter break homework, and I just don't remember how to do these, so can guys also explain how you got the answer? THANKS SO MUCH

Answer 1

Question 1
Let length = x m
Breadth = 70 - x m
A(x) = (x) (70 - x)
A(x) = 70x - x²
A `(x) = - 2x + 70 = 0 for max Area
x = 35
Length = 35 m
Breadth = 35 m

Could also obtain solution by completing the square:-
A(x) = - [ x ² - 70 x ]
A(x) = - [ (x ² - 70 x + 1225) - 1225 ]
A(x) = - [ (x - 35) ² - 1225 ]
x = 35 to give maximum value for A (as above)

Question 2
x + (x + 1) + (x + 2) = 378
3x + 3 = 378
3x = 375
x = 125
Numbers are 125 , 126 , 127

Answer 2

The key to answering your first question is knowing that the largest enclosed area would be made by forming a square. No combination of sides will generate more area than four sides of 35. This is known because the area of the quadrilateral is length times width, and we know that equal factors (i.e. sides of the same length) generate larger products (areas) than their unequal counterparts. For example, we know that a given number times itself will always be larger than two fewer than that number times two greater than the number.

x^2 > (x-2)(x+2)
x^2 > x^2 - 2x + 2x -4
x^2 > x^2 -4

Thus, it is clear that the dimensions should be 35 x 35, but I do not know how one arrives at that answer without first knowing the enclosure must be a square of four equal sides. This being the case, it strikes me that I have the right answer, but I'm not sure I'm showing you the proper algebraic way to solve it.

The second question is easier.

Let x = the lowest number of the three consecutive integers
Let x+1 = the second number in the series
Let x+2 = the third number in the series

x + (x+1) + (x+2) = 378
3x + 3 = 378
3x = 375
x = 125

Thus, the lowest number is 125, making the next numbers 126 and 127.

Good luck with your math!

Answer 3

the second one is x+(x+1)+(x+2)=378, then add up all the x's so the problem should be 3x+3=378, subtract 3 so the question is 3x=375, divide by 3 and the answer should be x=125 then plug in 125 in the problem and the answer should be 125,126, and 127

Answer 4

im not sure how to do the first one...but this is the second one...

x+x+1+x+2=378.....you set it up like this because x is the first integer and if you add one to x then you have the next consecutive integer and add 2 to get the next integer

hope this helps!

Answer 5

35x35

X+(X+1)+(x+2) = 378
3x+3=378
3x=375
x=125
125,126,127

Answer 6

The first one is 35m x 35m and the second one is 125,126,127

Answer 7

second one is 125,126,and 127

oh and im not sure how i got it, my mind just thinks that way