A rectangular garden is bounded on one side by a house and is to be fenced on the other three sides. Fencing material costs $16 per foot for the side parallel to the house and $20 per foot for the other two sides. What are the dimensions of the garden of largest possible area if $1600 is to be spent for fencing material?

Let the side parallel to the house is x. The other side is y. Total perimeter to be fenced is the three sides = x + y + y = x +2y Cost of fencing side which is y foot is 20*y Cost of fencing side parallel to house( x foot) = 16*x Total cost is 2* (20y) + 16x Total cost = 40y +16x. Cost = 1600 1600 = 40y +16x -------------(i) Divide the equation (i) by 4 400 = 10y + 4x (400 - 10y)/4 = x ------------(ii) Area = x*y. We want to maximize area Plug in value of x from (ii) (400-10y)/4 * y = area 100y - 5/2 y^2 = area Differentiating with respect to y, we get 100 - 5y =A'(y) 100 -5y = 0 , solving we get y = 20. x = (400 -10y)/4 = (400 -200)/4 = 200/4 = 50 Dimensions are 50 ft and 100 ft

Rita you made a mistake in your final answer. The math is there though. Consider the $1600 budget: \[2(20*50)+16(100)=3600\] If you plug the 20 in you solved for y: \[1600=40(20)+16x\] \[1600=800+16x\] \[800=16x\] \[x=50\] So two 20 ft runs and a single 50 ft run to get the maximum area. Like I said you did all the hard lifting :)

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