Question 1178856

The top and base of a fish tank are
rectangles whose length is 10
inches more than the width. 
let width be x
and length be y, y= x+10

sum of height and width of the
tank is equal to 50 inches
let z be the height of the tank
z+x+20=50 inches
z+x =30 

combined area,
2 * x(x+10) = 4zy-400

divide by 2
x^2+10x =2zy-200,   but y= x+10

----> x^2+10x=2(30-x)(x+10)
---->
 x^2+10x = 2(30x+300 -x^2-10x)
x^2 +10x = 60x +600 -2x^2-20x)
3x^2-30x -600=0
/3
x^2-10x-200=0
x^2-20x+10x-200=0

x(x-20)+10(x-20)=0
(x-20)(x+10)-0

Ignore negative x =20 width
length = x+10 = 30
z = height = 10





If the
, and the
combined area, of the top and the
base is 400 inches2
less than the
total area of four sides, what are
the dimensions of the tank?