can you answer these questions? its problem solving. its like make a table or sumthing.
1. you can cut square corners off an 11 in. by 14 in. piece of cardboard to get a pattern that you could fold into a box without a top.
a. what dimensions for the corners, to the nearest quater-inch, will give the greatest volume?
b. what is the greatest volume of the box to the nearest tenth?
2. corinda has 400 ft. of fencing to make a play area. she wants the fenced area to be rectangular. what dimensions should she use in order to enclose the maximum possible area?
3. the consecutive even intergers from 2 to n are 2, 4, 6, ..... n. the square of the sum of the intergers is 5,184. what is the value of n?
1. you can cut square corners off an 11 in. by 14 in. piece of cardboard to get a pattern that you could fold into a box without a top.
a. what dimensions for the corners, to the nearest quater-inch, will give the greatest volume?
b. what is the greatest volume of the box to the nearest tenth?
2. corinda has 400 ft. of fencing to make a play area. she wants the fenced area to be rectangular. what dimensions should she use in order to enclose the maximum possible area?
3. the consecutive even intergers from 2 to n are 2, 4, 6, ..... n. the square of the sum of the intergers is 5,184. what is the value of n?
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1) If a square of side length "d" is cut from each corner of your cardboard, the resulting box dimensions will be
.. volume = (11 - 2d)(14 - 2d)d = 4d^3 - 50d^2 + 154d
The derivative of this with respect to d is
.. d(volume)/dd = 12d^2 - 100d + 154 = 2(6d^2 - 50d + 77)
This has a root of interest at
.. d = (50 - √(50^2-4(6)(77))/(2(6)) = (25-√163)/6 ≈ 2.03
Cutting a 2" square from each corner will maximize the volume. That volume will be 7*10*2 = 140 in^3.
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If you are to solve this by making a table, you can choose corner square dimensions of 1", 2", and 3" to see what the general shape of the curve of volume versus square size will be. You will get 108 in^3, 140 in^3, and 120 in^3, respectively, for these choices. This suggests that your maximum will be between 2 and 3 inches. At 2.5", you get a volume of 135 in^3, so now your maximum looks like it is between 2 and 2.5. Trying 2.25", you get a volume of 138.9 in^3. To make sure that 2.00" is the maximum, you can try 1.75" for the square dimension and you will discover the volume to be 137.8 in^3. In 6 trials, you have found the maximum to the desired rounding.
2) The rectangle with the largest area for a given perimeter is a square. Corinda should fence an area of 100' by 100'.
_____
You know that a rectangle with a zero length side has zero area, so the area gets larger as the side gets longer. Once you have a square, the shortest side can get no longer, so you suspect that a square will have maximum area. You can try a rectangle that is not a square to see if it is larger. 99 ft by 101 ft, for example, gives an area of 9999 ft^2, less than the 10,000 ft^2 of a 100' by 100' rectangle.
3) The sum of N integers is N(N+1)/2, so your sum of even numbers with n as the last will be 2(n/2)(n/2+1)/2 = (n/2)(n/2+1). You have said this is equal to √5184 = 72, Factors of 72 that differ by 1 are 8 and 9, so we have n/2 = 8. The value of n is 16.
.. volume = (11 - 2d)(14 - 2d)d = 4d^3 - 50d^2 + 154d
The derivative of this with respect to d is
.. d(volume)/dd = 12d^2 - 100d + 154 = 2(6d^2 - 50d + 77)
This has a root of interest at
.. d = (50 - √(50^2-4(6)(77))/(2(6)) = (25-√163)/6 ≈ 2.03
Cutting a 2" square from each corner will maximize the volume. That volume will be 7*10*2 = 140 in^3.
_____
If you are to solve this by making a table, you can choose corner square dimensions of 1", 2", and 3" to see what the general shape of the curve of volume versus square size will be. You will get 108 in^3, 140 in^3, and 120 in^3, respectively, for these choices. This suggests that your maximum will be between 2 and 3 inches. At 2.5", you get a volume of 135 in^3, so now your maximum looks like it is between 2 and 2.5. Trying 2.25", you get a volume of 138.9 in^3. To make sure that 2.00" is the maximum, you can try 1.75" for the square dimension and you will discover the volume to be 137.8 in^3. In 6 trials, you have found the maximum to the desired rounding.
2) The rectangle with the largest area for a given perimeter is a square. Corinda should fence an area of 100' by 100'.
_____
You know that a rectangle with a zero length side has zero area, so the area gets larger as the side gets longer. Once you have a square, the shortest side can get no longer, so you suspect that a square will have maximum area. You can try a rectangle that is not a square to see if it is larger. 99 ft by 101 ft, for example, gives an area of 9999 ft^2, less than the 10,000 ft^2 of a 100' by 100' rectangle.
3) The sum of N integers is N(N+1)/2, so your sum of even numbers with n as the last will be 2(n/2)(n/2+1)/2 = (n/2)(n/2+1). You have said this is equal to √5184 = 72, Factors of 72 that differ by 1 are 8 and 9, so we have n/2 = 8. The value of n is 16.