Describe the level surfaces of the function. f(x,y,z)=x+3y+5z

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Describe the level surfaces of the function.  f(x,y,z)=x+3y+5z

mylouscrapza · Accepted Answer

f(x,y,z)=x+3y+5z&amp;nbsp;k is constant kx+3y+5z=k&amp;nbsp;Equation of a plane is&amp;nbsp;ax+by+cz=d, with a surface normal of&amp;nbsp;&amp;lt;a,b,c&amp;gt;.&amp;nbsp;Varying l wi;; result in different parallel planes. So f describes a familly of parallel planes.&amp;nbsp;

Eliza Beth13 · Accepted Answer

Answer:x+3y+5z−c=0Explanation:To begin, we can set up the equation x+3y+5z=c. This equation represents a plane in three-dimensional space. The level surfaces of the function are precisely these planes for different values of c.Now, let&#039;s rewrite the equation in standard form to get a clearer picture of the level surfaces. We rearrange the terms to obtain x+3y+5z−c=0. This form allows us to identify the coefficients of x, y, and z as 1, 3, and 5, respectively.We can represent the equation as follows:x+3y+5z−c=0Thus, the level surfaces of the function f(x,y,z)=x+3y+5z are planes in three-dimensional space, given by the equation x+3y+5z−c=0, where c represents the constant value.

Don Sumner · Accepted Answer

Step 1:To describe the level surfaces of the function f(x,y,z)=x+3y+5z, we need to find the values of x, y, and z that satisfy the equation f(x,y,z)=k, where k is a constant.Let&#039;s start by setting up the equation:x+3y+5z=kStep 2:To visualize the level surfaces, we can rewrite the equation in terms of z as the dependent variable:z=k−x−3y5This equation represents a plane with normal vector (1,3,5) passing through the point (x,y,k−x−3y5).The level surfaces of the function are therefore planes with varying values of k. Each plane is parallel to the plane defined by the equation x+3y+5z=0, which passes through the origin.By varying the value of k, we can obtain different parallel planes. When k=0, the plane passes through the origin and represents the level surface corresponding to the value k=0. For other values of k, the plane will be shifted parallel to the original plane.In summary, the level surfaces of the function f(x,y,z)=x+3y+5z are a family of parallel planes with varying values of k, all of which are parallel to the plane defined by the equation x+3y+5z=0.

madeleinejames20 · Accepted Answer

By setting f(x,y,z) equal to c, we can find the equations of the level surfaces.x+3y+5z=cThis equation represents a plane in three-dimensional space. For each value of c, there is a corresponding level surface that is a plane parallel to the original plane. The value of c determines the distance of the plane from the origin.By rearranging the equation, we can express it in the general form of a plane equation:x+3y+5z−c=0In this form, the coefficients of x, y, and z represent the normal vector to the plane.Thus, the level surfaces of the function f(x,y,z)=x+3y+5z are planes with equations of the form x+3y+5z−c=0, where c is a constant.

Describe the level surfaces of the function. f(x,y,z)=x+3y+5z

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