SAT Question 5b2b8866: Area and Volume answer and explanation with Desmos solution

4$.I{t}^{\prime }sgiventhattheareaoftheoriginalposteris＄360$ square inches. Let $l$ represent the length, in inches, of the original poster, and let $w$ represent the width, in inches, of the original poster. Since the area of a rectangle is equal to its length times its width, it follows that ＄360 = l w$.I{t}^{\prime }salsogiventhatacopyoftheposterismadeinwhichthelengthandwidthoftheoriginalposterareeachincreasedby＄20\%$. It follows that the length of the copy is the length of the original poster plus ＄20 %$ofthelengthoftheoriginalposter,whichisequivalentto$l + \frac{20}{100} l$inches.Thislengthcanberewrittenas$l + 0.2 l$inches,or＄1.2l$ inches. Similarly, the width of the copy is the width of the original poster plus ＄20 %$ofthewidthoftheoriginalposter,whichisequivalentto$w + \frac{20}{100} w$inches.Thiswidthcanberewrittenas$w + 0.2 w$inches,or＄1.2w$ inches. Since the area of a rectangle is equal to its length times its width, it follows that the area, in square inches, of the copy is equal to $\left(1.2l\right)\left(1.2w\right)$, which can be rewritten as $\left(1.2\right)\left(1.2\right)\left(lw\right)$. Since ＄360 = l w$,thearea,insquareinches,ofthecopycanbefoundbysubstituting＄360$ for $lw$ in the expression $\left(1.2\right)\left(1.2\right)\left(lw\right)$, which yields $\left(1.2\right)\left(1.2\right)\left(360\right)$, or ＄518.4$.Therefore,theareaofthecopy,insquareinches,is＄518.4$.