A rectangle is thrice along as it is wide .it has its dimension increased by 20$. find $ change in its area?

1 Answer
Apr 18, 2018

Not sure what you mean by #20$#

Worked out to a point where you can finish it off.

Explanation:

#color(brown)("Determine the rate of change in area")#

Set the length of width as #w#
Set area as #a#

From the question we have:

#"length" color(white)("ddd")"width"#
#color(white)("d")3wcolor(white)("ddddddd") w#

#a_1=(3w)xx(w) = 3w^2" "................Equation(1)#

Suppose we increase the width by the unknown value #delta#. Then we have:

#a_2=3(w+delta)xx(w+delta)#

#a_2=color(blue)((3w+3delta))xx color(green)((w+delta) )#

Multiply everything in the right bracket by everything in the left.

#a_2=color(blue)(3w)color(green)((w+delta) color(blue)(+3delta)color(green)((w+delta)))larr" Notice that the + follows "3delta #

#a_2=3w^2+3wdelta +3wdelta+3delta^2#

#a_2=3w^2+6wdelta +3delta^2" "........................Equation(2)#

To determine the change in area subtract #Eqn(1)" from "Eqn(2)#

#a_2=3w^2+6wdelta +3delta^2#
#ul(a_1= 3w^2 color(white)("ddddddddddd")larr" Subtract")#
#a_2-a_1=color(white)("d")6wdelta+3delta^2#

Rate of change is #(a_2-a_1)/delta# so divide both sides by #delta#

#(a_2-a_1)/delta = 6w+3delta#

Actually this is getting very close to calculus so it should be written as:

#(deltaa)/(deltaw)= 6w+3(deltaw)#

Where #delta" something"# measn a very little bit of 'something'.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Complying with your question content")#

Set #delta = 20$# giving

#(a_2-a_1)=6w(20$)+3(20$)^2#

#(a_2-a_1)=120w$+1200$^2#

#1200$^2+120$-(a_2-a_1)=0#

Using the quadratic formula:

#$=(-120+-sqrt(120^2-4(1200)(-(a_2-a_1))))/(2(1200))#

I will let you simplify this further.