#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'.
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#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.