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Chapter 12: Problem 78

Find an equation of the plane passing through the point (3,2,1) that slices off the solid in the first octant with the least volume.

Short Answer

Expert verified
The equation of the plane is: x + y + z = 6.

Step by step solution

01

Write the plane equation in intercept form

The intercept form of a plane equation is given by: \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] where (a, b, c) are the x, y, and z intercepts respectively. We need to find the equation of the plane that minimizes the volume of the solid in the first octant, which means we must minimize the triangle area formed by the intercepts in the xy-plane.
02

Find the area of the triangle formed by the intercepts in the xy-plane

The normal vector of the plane is proportional to the volume of the solid. Therefore, we can calculate the area of the triangle in the xy-plane formed by the intercepts as: \[A = \frac{1}{2}ab\] We need to minimize this area. By the Cauchy-Schwarz inequality we have: \[A^2 = \frac{1}{4}a^2b^2 \ge \frac{1}{3}(a^2 + b^2 + a^2 + b^2)\] Solve for ab: \[ab \ge \sqrt{3(a^2 + b^2)}\]
03

Calculate the equation of the normal vector and plane

Since (3, 2, 1) is a point in the plane, we can use the inequalities found in step 2 to write: \[9a + 8b = c\sqrt{3(a^2 + b^2)}\] Now, we can substitute (3, 2, 1) into the plane equation: \[\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\] Solve the system of equations: \[9a + 8b = c\sqrt{3(a^2 + b^2)}\] \[\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\] From these equations, we can find that a = 6, b = 6, and c = 6. So, the equation of the plane is: \[\frac{x}{6}+\frac{y}{6}+\frac{z}{6}=1\]
04

Final Answer

Thus, the equation of the plane passing through the point (3,2,1) that slices off the solid in the first octant with the least volume is: \[x+y+z=6\]

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Most popular questions from this chapter

Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the x y-, x z^{-}, \text {and } y z-\text {traces, when they exist. c. Sketch a graph of the surface. $$-\frac{x^{2}}{6}-24 y^{2}+\frac{z^{2}}{24}-6=0$$

Use the definition of differentiability to prove that the following functions are differentiable at \((0,0) .\) You must produce functions \(\varepsilon_{1}\) and \(\varepsilon_{2}\) with the required properties. $$f(x, y)=x+y$$

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May 1985, and Calculus and Analytical Geometry, 2nd ed., Philip Gillett, 1984)
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