# Optimization Problems with Functions of Two Variables

Several optimization problems are solved and detailed solutions are presented. These problems involve optimizing functions in two variables using first and second order partial derivatives.

## Problems with Detailed Solutions## Problem 1You decide to build a box that has the shape of a rectangular prism with a volume of 1000 cubic centimeters. Find the dimensions x, y and z of the box so that the total surface area of all 6 faces of the box is minimum.Solution to Problem 1:The total area A of all six faces of the prism is given by. A = 2 xy + 2 yz + 2 zx The volume of the box is given; hence xyz = 1000 Solve the above for z. z = 1000 / (xy) Substitute z in the expression of the area A to obtain. A(x,y) = 2xy + 2y * 1000 / (xy) + 2x * 1000 / (xy) = 2xy + 2000 / x + 2000 / y We now need to find x and y that minimize the area A. We first need to find the critical points and then test the second partial derivatives. The first order partial derivatives of A are given by A _{x}(x,y) = 2y - 2000 / (x^{2})
A _{y}(x,y) = 2x - 2000 / (y^{2})
The critical points are found by setting A _{x}(x,y) = 0 and A_{y}(x,y) = 0 and solving the system obtained. Which gives2y - 2000 / (x ^{2}) = 0 which gives 2yx^{2} = 2000
2x - 2000 / (y ^{2}) = 0 which gives 2xy^{2} = 2000
Solve the above to obtain x = 10 and y = 10 We now need to find the second order partial derivatives. A _{xx}(x,y) = 4000/(x^{3})
A _{yy}(x,y) = 4000/(y^{3})
A _{xy}(x,y) = 2
We now need to test the values of A _{xx}, A_{yy} and A_{xy} at the point (10,10) in order to use the theorem on minima and maxima of functions with 2 variables._{xx}(10,10) A_{yy}(10,10) - A_{xy}^{2}(10,10) = 4 * 4 - 4 = 12
D is positive and A _{xx}(10,10) = 4 is positive and therefore the area A is minimum forx = 10 cm y = 10 cm z = 1000/(xy) = 10 cm.
## Problem 2Find the dimensions of a six-faced box that has the shape of a rectangular prism with the largest possible volume that you can make with 12 squared meters of cardboard.
_{xx}(√2,√2) V_{yy}(√2,√2) - V_{xy}^{2}(√2,√2) = 5/2
D is positive and V _{xx}(√2,√2) = -√2 is negative and therefore the volume V is maximum for
x = √2 meters y = √2 meters z = (6- xy) / (x + y) = √2 meters.
## Problem 3Find the distance from the point (1,2,-1) to the plane given by the equation x - y + z = 3.
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