Answer to Question #259058 in Calculus for Ale

Question #259058

From a thin piece of cardboard 10 in. by 10 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary.




A. 6.7 in. x 6.7 in. x 3.3 in.; 148.1 in³




B. 3.3 in. x 3.3 in. x 3.3 in.; 37 in³




C. 5 in. x 5 in. x 2.5 in.; 62.5 in³




D. 6.7 in. x 6.7 in. x 1.7 in.; 74.1 in³

1
Expert's answer
2021-11-02T18:29:17-0400

Let x=x= the side of cutted square corner. Then a volume of a box will e


V=V(x)=(102x)2x, 0x5V=V(x)=(10-2x)^2x, \ 0\leq x\leq5

Find the first derivative with respect to xx


V(x)=((102x)2x)V'(x)=((10-2x)^2x)'

=(102x)2+2(102x)(2)x=(10-2x)^2+2(10-2x)(-2)x

=(102x)(102x4x)=4(5x)(53x)=(10-2x)(10-2x-4x)=4(5-x)(5-3x)

Find the critical number(s)


V(x)=0=>4(5x)(53x)=0V'(x)=0=>4(5-x)(5-3x)=0

Find the critical number(s)


x1=53,x2=5x_1=\dfrac{5}{3}, x_2=5

V(0)=0V(0)=0

V(5)=0V(5)=0


V(53)=(102(53))2(53)=20002774.1V(\dfrac{5}{3})=(10-2(\dfrac{5}{3}))^2(\dfrac{5}{3})=\dfrac{2000}{27}\approx74.1

The function V(x)V(x) fas the absolute maximum on [0,5][0, 5] with value of 200027\dfrac{2000}{27} at x=53.x=\dfrac{5}{3}.


531.7\dfrac{5}{3}\approx1.7


10253=2036.7(in)10-2\cdot\dfrac{5}{3}=\dfrac{20}{3}\approx6.7(in)

D. 6.7 in. x 6.7 in. x 1.7 in.; 74.1 in³




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment