Hi , i have this question i have answered the first two parts , but im not sure if there right and im also unsure about the last parts. Thank you for your help.

A piece of wire 16cm long is cut into two pieces . One piece is 8x cm long and is bent to form a rectangle measuring 3x cm by x cm long. the other piece is bent to form a square.
i) find in terms of x,
a) the length of a side of the square, i got the width to be 2cm and the length 4cm.
b)the area of the square , i got to be 4cm^2.

ii) Show that the combined area of the rectangle and the square is Acm^2, where A=7cm^2-16x+16. Not sure how to do this at all, not sure if i got the ones i worked out right :S

iii) lastly i need to find the value of x for which A has its mimimum value, and also find this minimum value for A.

2. Originally Posted by Chez_
Hi , i have this question i have answered the first two parts , but im not sure if there right and im also unsure about the last parts. Thank you for your help.

A piece of wire 16cm long is cut into two pieces . One piece is 8x cm long and is bent to form a rectangle measuring 3x cm by x cm long. the other piece is bent to form a square.
i) find in terms of x,
a) the length of a side of the square, i got the width to be 2cm and the length 4cm.
You know that 8x of the wire is used for the rectangle. The square is made of whatever is leftover, so the square has (16-8x) cm of wire. Since a square has 4 equal sides, the total length of the square divided by 4 will tell you the length of a single side of the square. so the length of 1 side is:
$\frac{16-8x}4 = (4-2x) cm$
Originally Posted by Chez_
b)the area of the square , i got to be 4cm^2.
Square the side
$(4-2x)^2 = (16-16x+4x^2)cm^2$
Originally Posted by Chez_
ii) Show that the combined area of the rectangle and the square is Acm^2, where A=7cm^2-16x+16. Not sure how to do this at all, not sure if i got the ones i worked out right :S
We already know the area of the square, now we need to find the area of the rectangle. We get this by multiplying the two sides together. They gave you the two sides as 3x and x, So
$3x*x = (3x^2)cm^2$

Now just add the two areas together
$(3x^2) + (16-16x+4x^2) = 16-16x+7x^2$

Originally Posted by Chez_
iii) lastly i need to find the value of x for which A has its mimimum value, and also find this minimum value for A.
So we have our formula for area
$A = 16-16x+7x^2$

Now, if we were to graph this, we would see that it has a point where it is minimized, meaning where the area is least. Because it is the least, the points around it must be greater than it, so this point is where the graph stops going down and begins going up. At this point, the tangent line is zero, meaning the slope is zero.

You know, then, that if you can find the equation of the slope, you can find out where it equals zero, and this will potentially be the point you are looking for.

We can find the equation of the slope by differentiating, so
$A\prime = \frac d{dx} 16-16x+7x^2 = -16+14x$

Now, we want to find out what x value will cause A' to be equal to zero, so we set it equal to zero and solve for x.
$0 = -16+14x$
$16 = 14x$
$\frac 87 = x$

Now, we have x = 8/7 as a point where our graph has a slope of zero, this means it might be the minimum, so we can check it in any of three ways. First is to plug it's value into our formula for area, then check a point on either side, and make sure it is less than those points, the second way is to check points on the derivative to make sure that it is decreasing to the point, and then increasing from the point (check a point before it should have a negative value, check a point after, it should have a positive value), and the third way is to take the derivative of the derivative (Area double prime) and plug the value into it's formula, if it is positive, then the derivative is increasing, so it will be a minimum, if it is negative, it will be a maximum.

I'll just plug it into our initial formula:
$A(1) = 7$
$A(8/7) = 6.85$
$A(9/7) = 7$

So we can see that the points around it are greater than it, which means it is a minimum value. Now we just need to check our endpoints, since they will not necessarily have tangents equal to zero, but could still be less than our value (not really in this problem, but it's good to get into the practice of checking, for other problems)

Obviously the lowest value x can be is zero, in which case the entire wire is used for the square. And the highest it can be is 2, as 2*8=16, and thus the entire wire is used for the rectangle. So we check these endpoints
$A(0) = 16$
$A(2) = 12$

So this means that these points are not less than x=8/7, so then x=8/7 must be the minimum value.