a. Length of one side of second square is 10-x/4 and first must be x/4. There's 10 cm total so you can construct the first square with perimeter "x" cm and therefore the side length must be x/4 since all sides are equal in a square. The remaining wire can be expressed as Total WIre - Used Wire so "10-x" and side lengths are equal in a square once again being 10-x/4.
tysm!!!!a. Length of one side of second square is 10-x/4 and first must be x/4. There's 10 cm total so you can construct the first square with perimeter "x" cm and therefore the side length must be x/4 since all sides are equal in a square. The remaining wire can be expressed as Total WIre - Used Wire so "10-x" and side lengths are equal in a square once again being 10-x/4.
b. side length of the first square is x/4 and second is 10-x/4. To find area of both just;
...... (algebra)
c.
now find the minima. Stationary points have gradient of 0 so;
hence stat point at x = 5
second derivative tells us that the function is concave up for all values meaning the stat point must be a minima.
d. minimised area
substitute x = 5 into the total Area formula from (b) and you should get the value which should be 25/8 cm^2?
ok ok tysmmmmma) The side length of each square now how many sides are equal in a square? If you want the side divide by 4 from the total length used and the same applied to the other one I think that should be 10-x divided by 4 for the whole thing. If we did it your way it would become 40-x if we add all the lengths up which is not what we want. I really think ExtremelyBored User got so bored with maths that he misread the question.
A = (x/4)^2 + (10-x/4)^2 I think you can say A = (x/4)^2 + ((10-x)/4)^2a. Length of one side of second square is 10-x/4 and first must be x/4. There's 10 cm total so you can construct the first square with perimeter "x" cm and therefore the side length must be x/4 since all sides are equal in a square. The remaining wire can be expressed as Total WIre - Used Wire so "10-x" and side lengths are equal in a square once again being 10-x/4.
b. side length of the first square is x/4 and second is 10-x/4. To find area of both just do the area formula for both squares (side * side);
A = (x/4)^2 + (10-x/4)^2
...... (algebra)
c. dA/dx:
x-5/4
now find the minima. Stationary points have gradient of 0 so;
dA/dx = 0
hence stat point at x = 5
second derivative tells us that the function is concave up for all values meaning the stat point must be a minima.
d. minimised area
substitute x = 5 into the total Area formula from (b) and you should get the value which should be 25/8 cm^2
Yes that is what I meant.. There are four sides in a square and total cm for first square is "x" so finding the side length would be simply x/4 since x represents the total length of the square. Apologies if that was unclear.a) The side length of each square now how many sides are equal in a square? If you want the side divide by 4 from the total length used and the same applied to the other one I think that should be 10-x divided by 4 for the whole thing. If we did it your way it would become 40-x if we add all the lengths up which is not what we want. I really think ExtremelyBored User got so bored with maths that he misread the question.