(a) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) Height, x Length and Width Volume, V 36 - 2(1) 1(36 – 2(1)]² = 1156 36 - 2(2) 2[36 - 2(2)]2 = 2048 3 36 - 2(3) 3[36 - 2(3)]² = 36 – 2(4) 4[36 - 2(4)]² = 36 - 2(5) 5[36 - 2(5)]² = 36 – 2(6) 6[36 – 2(6)]² = 6 Use the table to guess the maximum volume. V = (b) Write the volume V as a function of x. V = , 0

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.4: Complex And Rational Zeros Of Polynomials
Problem 39E
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An open box of maximum volume is to be made from a square plece of material, s = 36 inches on a side, by cutting equal squares from the corners and turning up the sides (see figure).
(a) Analytically complete six rows of a table such as the one below. (The first two rows are shown.)
Length and
Width
Height, x
Volume, V
36 - 2(1)
1(36 - 2(1)]2 =
= 1156
2
36 - 2(2)
2[36 - 2(2)]2 = 2048
36 – 2(3) 3[36 - 2(3)]2 =
3
36 – 2(4) 4[36 - 2(4)]² :
4
36 – 2(5) 5[36 - 2(5)]? :
36 – 2(6) 6[36 - 2(6)]2 =
Use the table to guess the maximum volume.
V =
(b) Write the volume V as a function of x.
V =
0 < x < 18
(c) Use calculus to find the critical number of the function in part (b) and find the maximum value.
V =
S - 2r
Transcribed Image Text:An open box of maximum volume is to be made from a square plece of material, s = 36 inches on a side, by cutting equal squares from the corners and turning up the sides (see figure). (a) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) Length and Width Height, x Volume, V 36 - 2(1) 1(36 - 2(1)]2 = = 1156 2 36 - 2(2) 2[36 - 2(2)]2 = 2048 36 – 2(3) 3[36 - 2(3)]2 = 3 36 – 2(4) 4[36 - 2(4)]² : 4 36 – 2(5) 5[36 - 2(5)]? : 36 – 2(6) 6[36 - 2(6)]2 = Use the table to guess the maximum volume. V = (b) Write the volume V as a function of x. V = 0 < x < 18 (c) Use calculus to find the critical number of the function in part (b) and find the maximum value. V = S - 2r
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