Question

Animal Control * e: Optimization. skille Emphasized: Derivative. Nearly al1 calculus textbooks contain fencing problems. This problem puts that type of problem in a real-life context. Farms often contain various types of animals which need to be . Suppose a farm has bulls, steers, calves, heifers, milk-cows with calves, and cows Just for milking. Bulls aren't near the heifers because it riles them up, cows which have calves have their milk reserved for their babies. Colostrum is the first milk produced by the cows after they have their babies. It's some of the most nutrient dense milk you can find. It's filled with beneficial imnunoglobulins, growth factors, and antibodies that promote a healthy immune system, and also oligosaccharides that promote healthy bacteria growth in the calves' digestive system. A pen configuration that accommodates these criteria separated by individual pens and could look like this. A farmer has $10,560 to construct the pens as shown. The four sides of the Bull's pen costs $12 per linear foot other fence lengths cost $8 per linear foot and all the dimensions of each animal's pen if we want to maximize the total area of the penned region? 2 0 Caes X Heifers X Stees 2 Buls XCows WX Cows wo Calves 2X Calves

Answer 1

A r haos-to Gsnstruct the Pens aus shoum below And the tour stdes b the bulfs pen Costs a 12 here we ee to tjnd out dimenstons eah anrals pen rm tb,560. nce e. Lue ne ue want to moninnze the-total area Buls 2x the Qiver cdata> we are tonting an equaion) that tohee 12,8, lo56o are given value* uta, 2У.14 are the pens loc trd values Buls above and lvoer Values Au x's except 2x are equal to x (20χ+3 ) 10,560 20 x tay. to 560

But , -the avea) is given b 6x 320 20 1320-2 DX 3 Substitttir? y Value to eqn 3 3 3 3 2. 3

dt dt 3 3 3 ul Substituting Value ; 1320- 20 (6o) 3 The dinensiơns。each.anl mals Pos pers and Gar
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