Chapter 4 - Applications of the Derivative - 4.4 Optimization Problems - 4.4 Exercises - Page 274: 6

$${\text{Width}} = {\text{length}} = \frac{1}{4}P$$

$$\eqalign{ & {\text{From the image shown below we have that the perimeter is}} \cr & {\text{given by the equation}} \cr & 2x + 2y = P{\text{ }}\left( {\bf{1}} \right) \cr & {\text{In this case}},{\text{ the objective function is the area }}A = xy \cr & A = xy{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}y \cr & 2y = P - 2x \cr & {\text{ }}y = \frac{1}{2}P - x \cr & {\text{Substitute }}\frac{1}{2}P - x{\text{ into the equation }}\left( {\bf{2}} \right) \cr & A = x\left( {\frac{1}{2}P - x} \right) \cr & {\text{Which is a function of the single variable }}x.{\text{ Notice that the }} \cr & {\text{values of }}x{\text{ lie in the interval }}0 \leqslant x \leqslant \frac{1}{2}P{\text{ with }}A\left( 0 \right) = A\left( {\frac{1}{2}P} \right) = 0 \cr & {\text{Differentiate }}A{\text{ with respect to }}x \cr & A = \frac{1}{2}Px - {x^2} \cr & \frac{{dA}}{{dx}} = \frac{1}{2}P - 2x \cr & {\text{Find the critical points by solving }}\frac{{dA}}{{dx}} = 0 \cr & \frac{1}{2}P - 2x = 0 \cr & x = \frac{1}{4}P \cr & {\text{To find the absolute maximum value of }}A{\text{ on the interval }}\left[ {0,\frac{1}{2}P} \right]{\text{ }} \cr & {\text{we check the endpoints and the critical points}}.{\text{ Because }} \cr & A\left( 0 \right) = A\left( {\frac{1}{2}P} \right) = 0{\text{ and }}A\left( {\frac{1}{4}P} \right) = \frac{1}{{16}}{P^2} > 0,{\text{ }} \cr & {\text{we conclude that }}A{\text{ has its absolute maximum value at }}x = \frac{1}{4}P \cr & {\text{Now we can calculate }}y \cr & {\text{ }}y = \frac{1}{2}P - x \cr & {\text{ }}y = \frac{1}{2}P - \frac{1}{4}P \cr & {\text{ }}y = \frac{1}{4}P \cr & {\text{Therefore, the dimensions are Width}} = {\text{length}} = \frac{1}{4}P \cr} $$