Question 1178856: The top and base of a fish tank are
rectangles whose length is 10
inches more than the width. If the
sum of height and width of the
tank is equal to 50 inches, 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?
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
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?
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