Find the local maximum and minimum values and saddle point(s) of the function. I

Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x,y)=2x^3+xy^2+5x^2+y^2+9 local maximum value(s) local minimum value(s) saddle point(s) (x, y, f) = SEssCalcET2 11.7.034. Find the points on the surface y^2 = 1 +xz that are closest to the origin. (x, y, z) = () (smaller y-value) (x, y, z) = () (larger y-value) SEssCalcET2 11.7.040. Find the dimensions of the rectangular box with largest volume if the total surface area is given as 100 cm^2. (Let x, y, and z be the dimensions of the rectangular box.) (x, y, z) = ( )

Explanation / Answer

f(x, y) = 2x^3 + xy^2 + 5x^2 + y^2 + 1

Partial Derivatives
f_x = 6x^2 + y^2 + 10x
f_y = 2xy + 2y

When at critical points, f_x = 0, f_y = 0

6x^2 + y^2 + 10x = 0
==> y^2 = -6x^2 - 10x ............ [1]

2xy + 2y = 0
==> y(x + 1) = 0
==> y = 0 and/or x = -1

But y = 0 and x = -1 cannot be a solution because it doesn't satisfy [1]

When y = 0, from [1]:
-6x^2 - 10x = 0 ==> x = 0, -5/3

When x = -1
y^2 = -6(-1)^2 - 10(-1) = 4 ==> y = -2, 2

Therefore the critical points are (0, 0), (-5/3, 0), (-1, -2), (-1, 2)

Partial Derivatives again
f_xx = 12x + 10
f_xy = 2y
f_yy = 2x + 2

Second Derivative Test
D(x, y) = f_xx (x, y) * f_yy (x, y) - [ f_xy (x, y) ]^2

(0, 0) : D(0, 0) = 10 * 2 - 0^2 = 20 > 0, f_xx (0, 0) > 0, Local Minimum
(-5/3, 0) : D(-5/3, 0) = -10 * -4/3 - 0^2 = 40/3 > 0, f_xx (-5/3, 0) > 0, Local Minimum
(-1, -2) : D(-1, -2) = -2 * 0 - 4^2 < 0, Saddle Point
(-1, 2) : D(-1, 2) = -2 * 0 - 4^2 < 0, Saddle Point

So (0, 0) and (-5/3, 0) are Local Mimimums; (-1, -2) and (-1, 2) are Saddle Points

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