Question

7 At Noon ship to A is 150km want to shup B. Ship a is se sasling east to 35km/h And Ship B is Salling noith at 25km/h iHow fast is the distance between the two steps. Changing at 4pm. - I A farmor with goofect of fencing want to enclose a rectanguler area and then divided it into three Penap Pery with bencing paralled to one side of the rectangle. what are the dimansions that will matmore the total area of the three Pene?

Answer 1

A n isokoni B a het after this A corea distince niken and D B coreas distance y kom » Aan BB'=y. from abone bisure distance between ships & A'B' = l kon lei) By py Nanayoors theoren A'B'? - ÁB? + BB!" ľ= (Iso-B)~ 7 y d= [ (150-n)?tyr non ditleren hare with resped bt e d [ (iso-95 ty]* + [lso ) + sgk. (190-19:36] + [150-15 +09 [16:50-) (***)*2) Viso-whtyn (n-150) dan ry day - 0 trefoe foot a t /+ 25 A x Incoleasing a = speed on A=35 d - Speed on B = 25 we need to atter uby on sailies.

n=35+4= iso, Pustig allvals in y=25*4= loo lo . (140-150x35. , (iso-140 +25210 یلدا ( ۱- - - - Kon/20 so dis konce between skrips decreasing at a Speed on 10 koulho at ap.on. o bencin was made posallel to side or least nfl other side or rechnules croce = 307 To ke area of rechinsular bield A = nx 38 = 3ny - Tokal length or bencino 30x2 + nthaw (sive) 34 + 2n=30 aD 3y = 300-m

Putting value of 3y in C H = n(30-20) A Boon-an Now we have to maximize A. differentiale A. dana 3ro.A (2) - 2.4 (12) da = 300 - 2x2n. do 300-40. At coitical point dat zo 300-4000 Sn=300 2 * = 350-75. ditterentiale da ascin. ILA = 300.2 (1)-2. In - 32o - 44) ~240 ns de coal n-7590D A is maximized. Ir 300-2x75 = 300 - 15o = 150 250 so reckasle will be on dimension 37=1soge on sides 754 and 15044 to whese bencing have to be done parallel to sides of 75lt 3