Mathematics OpenStudy (anonymous):

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? OpenStudy (anonymous):

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 OpenStudy (anonymous):

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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