Answer:
x = 4
y = 11
Step-by-step explanation:
Set your formula as follow:
2x+3=3x-1 -> 2x+3-3x=-1 -> -x+3 = -1 -> -x=-4 -> x=4
now substitute x for 4 in the following
y-2=2(4)+1 -> y-2=8+1 ->y-2 = 9y -> y=11
The questions for this problem would be:
1. What is the dimensions of the box that has the maximum volume?
2. What is the maximum volume of the box?
Volume of a rectangular box = length x width x height
From the problem statement,
length = 12 - 2x
width = 9 - 2x
height = x
where x is the height of the box or the side of the equal squares from each corner and turning up the sides
V = (12-2x) (9-2x) (x)
V = (12 - 2x) (9x - 2x^2)
V = 108x - 24x^2 -18x^2 + 4x^3
V = 4x^3 - 42x^2 + 108x
To maximize the volume, we differentiate the expression of the volume and equate it to zero.
V = 4x^3 - 42x^2 + 108x
dV/dx = 12x^2 - 84x + 108
12x^2 - 84x + 108 = 0x^2 - 7x + 9 = 0
Solving for x,
x1 = 5.30 ; Volume = -11.872 (cannot be negative)
x2 = 1.70 ; Volume = 81.872
So, the answers are as follows:
1. What is the dimensions of the box that has the maximum volume?
length = 12 - 2x = 8.60
width = 9 - 2x = 5.60
height = x = 1.70
2. What is the maximum volume of the box?
Volume = 81.872
9514 1404 393
Answer:
OL = 7
Step-by-step explanation:
Triangle JKL is isosceles and N is the midpoint of JL, so NL = NJ = 14.
Triangle LMN is isosceles and O is the midpoint of NL, so NL = 2(OL).
OL = NL/2 = 14/2
OL = 7
__
There is no WY.
Three unit's would be the numbers:
1, 2 and 3.
Any number you choose must be greater than 3 for example:
5, 8 or 10
Hope it helped,
BioTeacher101
