   Chapter 14.5, Problem 26E Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Manufacturing Show that a box with a square base, an open top, and a fixed volume requires the least material to build if it has a height equal to one-half the length of one side of the base.

To determine

To prove: That a box has height equal to one half the length of one side of the base if the box has a square base, an open top, and fixed volume and required the least material to build.

Explanation

Given Information:

A box with a square base, an open top, and a fixed volume that requires the least material.

Formula used:

According to the Lagrange multipliers method to obtain maxima or minima for a function z=f(x,y) subject to the constraint g(x,y)=0,

Step 1: Find the critical values of f(x,y) using the new variable λ to form the objective function F(x,y,λ)=f(x,y)+λg(x,y).

Step 2: The critical points of f(x,y) are the critical values of F(x,y,λ) which satisfies g(x,y)=0.

Step 3: The critical points of F(x,y,λ) are the points that satisfy Fx=0, Fy=0, and Fλ=0, that is, the points which make all the partial derivatives of zero.

For a function f(x,y), the partial derivative of f with respect to y is calculated by taking the derivative of f(x,y) with respect to y and keeping the other variable x constant. The partial derivative of f with respect to y is denoted by fy.

Power of x rule for a real number n is such that, if f(x)=xn then f(x)=nxn1.

Constant function rule for a constant c is such that, if f(x)=c then f(x)=0.

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

The surface area of a box with open top and length l, width w, and height h is 2lh+2wh+lw.

The volume of a box with length l, width w, and height h is lwh.

Proof:

Recall that, the surface area of a box with open top and length l, width w, and height h is 2lh+2wh+lw.

Let x be the length of the side of the square base and y be the height of the box.

Thus, the surface area of a box with open top and length x, width x, and height h is 2xy+2xy+x2=4xy+x2.

Thus, minimize the function f(x,y)=4xy+x2.

Recall that, the volume of a box with length l, width w, and height h is lwh.

Since, x be the length of the side of the square base and y be the height of the box.

Thus, the volume of the box is given by xxy=x2y.

Let the fixed volume of the box be V.

Thus, x2y=V.

The provided constraint is x2y=V.

According to the Lagrange multipliers method,

The objective function is F(x,y,λ)=f(x,y)+λg(x,y).

Thus, f(x,y)=4xy+x2 and g(x,y)=x2yV

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