Question

[-/2 Points] DETAILS LARCALCET7 4.7.017. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A rectangular poster is to contain 288 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch! What should the dimensions of the poster be so that the least amount of poster is used? in (smaller value) in (larger value) Need Help? Read it Watch Talk to a Tutor [-/2 Points] DETAILS LARCALCET7 4.7.019. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A farmer plans to enclose a rectangular pasture adjacent to a river (see figure). The pasture must contain 5,000 square meters in order to provide enough grass for the herd. No fencing is needed along the river. What dimensions will require the least amount of fencing? m x = y = m

please i need help with this questions because i don't understand

Answer 1

Izinches klinch Print Area 288 288 (x-1-1) (9-2-2) = 288 (2-2)(7-4) + y = 2884 +4 Poster Area = xy 1288 X-2 +4 for minimum avea of postes & (Poster Area) da X-288 7 tof 288 x-2 +4JX13 (x-2)2 xlo + 288 +420 x-2 -288x (x-2)2 – 288x + 288 (2-2) +4(2-2) ?=0 -576 + 4(x - 2)2 - 0 1576 = 12 X-2 = 4

x = 14 in ya 14-2 288 28 in y 2444 dimension x=14in smaller value 4 -28% in larger volue let dimension is x and Area xy ay= fence x+2y =x+10000 for minimum amount of fencing d(fence) वर d da 5000 A ett * 10000 °) 1 - 10000 x² x = 100 m and ya 5000 som x = loom y = 150/m

length. 280 arca width - y porimeter 2(x+4) - y = 140-x Area ny = x (140) - 140x - x² maximum d(Area) dx -> 140-22.50 - x= 70 m y = 140-2=140-70 y-Four length = 170m Width = 170 m let point is (ag) on y=x6 distance - √(x-0)2 + (y + 4)² d = x² + (y + 4) = 24 x²6+42 for minimum distance d (d) -o a f (x² + (x - 2)²) da ) 2x + 2(222) x 2x = 0 = x=0 os 1+ 2(x2-2) CO > x= 3) = 3 SO

250, 13 y = x²65-6, 2-6 - 2/14 distance - x² + (7 +42 for (0,6) of 10²+(-6+432 2 - T + (-21 +42 3 4 = 137.32 Hence closest points 21 smalles a value ( 16 - 22 larger x value