X+y=16
the sum of the squares is
x^2+y^2=sum
solve for y in first equation
x+y=16
y=16-x
subsitiute that for y in other equation
x^2+(16-x)^2=sum
x^2+x^2-32x+256=sum
2x^2-32x+256=sum
take derivitive to find the minimum value (or just find the vertex because the parabola opens up)
derivitive is
4x-32=derivitve of sum
the max/min is where the derivitive equals 0
4x-32=0
4x=32
x=8
at x=8
so then y=16-8=8
the smallest value then is 8^2+8^2=64+64=128
Hopefully this helps
C=10
13.
1st: split the 1 shape into 2 shapes. (spit across to where you see one right angle(top), and a trapezoid(bottom)
2nd: now what i did was fill that gap to where the total height is 9 and the total length is 7. ( if you don't want to do that it is fine )
3rd: for the top do 5 times 2 = 10
4th: for the bottom do 4 times 5 = 20
5th: add 10 (top) and 20 (bottom) = 30
You can set this up as 2 fractions equal to each other, like ratios.
18/11=108/x
Cross multiply and you get 18x=1,188
Now divide both sides by 18 and get x.
x=66.
If the height of the building is 108ft, the width is 66ft.
