The length of a rectangle is given by 5t+4sqrt(t), and its height is sqrt(t), where t is time in seconds and the dimensions are in centimeters. Fine the rate of change of the area of the rectangle with respect to time when the length of the rectangle is 9 centimeters.

Answer:11.5 cm²/sStep-by-step explanation:We must solve two problems to answer this question:find the time at which the length is 9 cmfind the rate of change of area with respect to time (at that time)For the first problem, we have ...... 9 = 5t +4√t... 9 -5t = 4√t . . . . . subtract 5t... 81 -90t +25t² = 16t . . . . square both sides... 25t² -106t +81 = 0 . . . subtract 16t... (t -1)(25t -81) = 0 . . . . factor... t = 1 or 3.24 . . . . . . t=3.24 is an extraneous solution_____For the second problem, we have area (a(t)) is ...... a(t) = length×height = (5t +4√t)(√t) = 5t^(3/2) +4tThen the derivative is ...... a'(t) = (3/2)(5√t) +4and... a'(1) = (3/2)(5√1) +4 = 11.5 . . . . . cm²/s_____Comment on the attachmentA graphing calculator can help solve both problems. It can help find the time at which length is 9 cm by solving ... length-9 = 0. (The nice thing here is that there is no extraneous solution.) It can compute the derivative of a function, too.