Answer:
The values of x and y that will minimize the cost is 15 ft and 5 ft respectively and the minimum cost will 3,375
Step-by-step explanation:
With a 1125 ft^3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top lush with the ground to catch runoff water, x is the length, y is the height;
x^2*y = 1125
y = 1125/x^2 ----(1)
The total cost of c = 5(x^2 + 4xy) + 10xy
Then,
c = 5(x^2 + 4x*(1125/x^2)} + 10x*(1125/x^2)
c = 5x^2 + 22,500/x + 11,250/x
c = 5x^2 + 33,750/x
To minimize the cost of c, find the derivatives of c; c'
c' = 10x - 33,750/x^2
c' = (10x^3 - 33,750)/x^2
c'(x) = 0; 10x^3 = 33,750
x = ³√33,750 = 15
x = 15; y = 1125/x^2
y = 1125/15^2 = 5
(x,y) = (15,5)
c = 5x^2 + 33,750/x
c = 5*15^2 + 33,750/15 = 3,375
The values of x and y that will minimize it is 15 ft and 5 ft respectively and the minimum cost will 3,375
The values of x and y that will minimize the cost is 15ft and 5ft. The minimum cost is 3,375
We have volume of a rectangular tank = 1125 ft^3
Side of a rectangular tank = x ft and
Length of a rectangular tank =y ft
So, [tex]x^2y = 1125[/tex]
[tex]y = \frac{1125}{x^2} ...(1)[/tex]
The total cost of rectangular tank (c) = [tex]5(x^2 + 4xy) + 10xy[/tex]
[tex]c = 5(x^2 + 4x(\frac{1125}{x^2} ) )+ 10x(\frac{1125}{x^2} )\\c = 5x^2 +\frac{22,500}{x} + \frac{11250}{x} \\c = 5x^2 + \frac{33750}{x}[/tex]
Differentiating it w.r.t x, we get
[tex]c' = 10x - 33,750/x^2\\c' =\frac{10x^{3}-33750 }{x^{2} } \\c'(x) = 0\\10x^3 = 33,750\\x = \sqrt[3]{33,750} \\ = 15\\x = 15; y = \frac{1125}{x^2} \\y = \frac{1125}{15^{2} } \\= 5\\(x,y) = (15,5)\\c = 5x^2 + \frac{33750}{x} \\c = 5(15)^2 + 33,750/15 \\= 3,375[/tex]
The values of x and y that will minimize the cost is 15 ft and 5 ft and the the minimum cost is 3,375.
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