Questions: A 3125 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 flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy. a. If the total cost is c=6(x^2+4xy)+36xy, what values of x and y will minimize it? b. Give a possible scenario for the cost function in part (a). a. The values of x and y that minimize the cost are x= and y= The minimum cost is .

A 3125 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 flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.
a. If the total cost is c=6(x^2+4xy)+36xy, what values of x and y will minimize it?
b. Give a possible scenario for the cost function in part (a).
a. The values of x and y that minimize the cost are x= and y=
The minimum cost is .

Transcript text: A $3125 \mathrm{ft}^{3}$ open-top rectangular tank with a square base xft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy. a. If the total cost is $c=6\left(x^{2}+4 x y\right)+36 x y$, what values of $x$ and $y$ will minimize it? b. Give a possible scenario for the cost function in part (a). a. The values of $x$ and $y$ that minimize the cost are $x=$ $\square$ and $y=$ $\square$ $\square$ The minimum cost is $\$$ $\square$ .

Solution

Solution Steps

To minimize the cost function given the constraints, we need to use calculus, specifically the method of Lagrange multipliers or substitution. First, express $ y $ in terms of $ x $ using the volume constraint $ x^2y = 3125 $. Substitute this expression into the cost function to get a function of $ x $ only. Then, find the derivative of this function with respect to $ x $, set it to zero, and solve for $ x $. Finally, use the value of $ x $ to find $ y $ and calculate the minimum cost.

Step 1: Volume Constraint

The volume of the tank is given by the equation: \[ x^2 y = 3125 \] From this, we can express $ y $ in terms of $ x $: \[ y = \frac{3125}{x^2} \]

Step 2: Cost Function

The cost function is defined as: \[ c = 6\left(x^2 + 4xy\right) + 36xy \] Substituting $ y $ into the cost function gives: \[ c = 6x^2 + 187500/x \]

Step 3: Finding the Minimum Cost

To minimize the cost, we take the derivative of $ c $ with respect to $ x $: \[ \frac{dc}{dx} = 12x - \frac{187500}{x^2} \] Setting the derivative equal to zero to find critical points: \[ 12x - \frac{187500}{x^2} = 0 \] Solving this equation yields: \[ x = 25 \] The negative solutions are not physically meaningful in this context.

Step 4: Calculate $ y $ and Minimum Cost

Substituting $ x = 25 $ back into the equation for $ y $: \[ y = \frac{3125}{25^2} = 5 \] Now, substituting $ x $ and $ y $ into the cost function to find the minimum cost: \[ c = 6(25^2) + 187500/25 = 11250 \]