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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Find the maximum volume of a box that is resting on the xy-plane with one vertex at the origin and the opposite vertex in the plane 2 x +   4 y   + z =   8 .

To determine

To calculate: The maximum volume of a box that is resting on the xy plane with one vertex

at the origin and opposite vertex 2x+4y+z=8.

Explanation

Given Information:

A box is resting on the xy plane with one vertex at the origin and opposite vertex in the

plane 2x+4y+z=8

Formula used:

The volume of a cube is given as,

V=xyz

Where, x, y, and z are the sides of the cube respectively.

Volume of a cube of side x,y and z units is V=xyz

The second partial derivative test for a multivariate function f(x,y).

Equate the first partial derivative of the function to 0.

fx=0 and fy=0

To test the relative extrema of the function

Step 1: Find the second partial derivative of the function, fxx, fyy,and fxy

Step 2: Evaluate the quantity,

d=fxx(a,b)fyy(a,b)[fxy(a,b)]2

Step 3: Estimate the nature of d.

If the value of d is positive and fxx(a,b)>0, so the function has a relative minimum.

If the value of d is positive and fxx(a,b)<0, so the function has a relative maximum.

If the value of d is negative do, the function has a saddle point at (a,b).

Calculation:

Consider the sides of the cube is x, y, and z.

Since one vertex of the cube lying on a plane,

2x+4y+z=8

Rearrange the above equation in term of z.

z=82x4y

Use the formula for the volume of the cube,

V=xyz

Substitute 82x4y for z.

V=xy(82x4y)=8xy2x2y4xy2

Now, V becomes a function of x and y so it can be write as,

V(x,y)=8xy2x2y4xy2

Use the simple power rule to differentiate the provided function keeping y as a constant.

Vx=x(8xy2x2y4xy2)=8y4xy4y2

Equate the first derivative of function with respect to x to zero,

8y4xy4y2=0y(84x4y)=0

From the zero product of multiplication,

y=0

Or,

84x4y=0

Substitute 0 for y in above equation.

84x4(0)=04x=8x=2

So, (2,0) is a vertex point of the cube.

Substitute 0 for x in above equation.

84(0)4y=04y=8y=2

So, (0,2) is a vertex point of the cube.

Use the simple power rule to differentiate the provided function keeping x as a constant.

Vy=y(8xy2x2y4xy2)=8x2x28xy

Equate the first derivative of function with respect to y to zero,

8x2x28xy=0x(82x8y)=0

From the zero product of multiplication,

x=0

Or,

82x8y=0

Substitute 0 for x in above equation.

82(0)8y=0y=1

So, (0,1) is a vertex point of the cube

Substitute 0 for y in above equation.

82x8(0)=0x=4

So, (4,0) is a vertex point of the cube

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