Thread: word problem 2 (rectangular)

i have another question on this topic.

a rectangular building is to have an area of 1800 sq. meters. on each side of the building there must be a walkway of 5 meters and on each end a walkway of 10 meters. What is the dimensions of the smallest rectangular piece of land that can accommodate the building?

so, i tried
area = xy
= (x-5)(x-10)
1800 = (x-5)(x-10)
= x^2 - 15x +50
0 = x^2 - 15x - 1750

doesn't seem to be im doing correctly...
what should i do?

Originally Posted by haebinpark

i have another question on this topic.

a rectangular building is to have an area of 1800 sq. meters. on each side of the building there must be a walkway of 5 meters and on each end a walkway of 10 meters. What is the dimensions of the smallest rectangular piece of land that can accommodate the building?

so, i tried
area = xy
= (x-5)(x-10)
1800 = (x-5)(x-10)
= x^2 - 15x +50
0 = x^2 - 15x - 1750

doesn't seem to be im doing correctly...
what should i do?

let L = building length
W = building width

LW = 1800

land area = A ...

A = (L+20)(W+10)

use the first equation to solve for L in terms of W (or W in terms of L ... it doesn't make a difference) and substitute into the second equation to area in terms of a single variable.

take the derivative and minimize.

umm.. on A = (L+20)(W+10), here,
why L+20...?
where is 20 come from?

Originally Posted by haebinpark

umm.. on A = (L+20)(W+10), here,
why L+20...?
where is 20 come from?

10 meters of walkway on each side