Question

asked Oct 8, 2023 157k views

1 Answer

Aenw · Answer 1 · 2023-10-15T07:04:41+0000

Answer:

B. 'x' should be 1.53 inches

Explanation:

Given equation for the volume of the box:

V=(8-2x)(11-2x)(x)

To find the value of x that will maximize the volume, differentiate the volume, set the first derivative to zero and solve for x.

Expand the equation:

V=x(8-2x)(11-2x)

$\implies V=x(88-16x-22x+4x^2)$

$\implies V=4x^3-38x^2+88x$

Differentiate:

$\implies (dV)/(dx)=12x^2-76x+88$

Set to zero and solve for x:

$\implies (dV)/(dx)=0$

$\implies 12x^2-76x+88=0$

Divide by 4:

$\implies 3x^2-19x+22=0$

Use the quadratic formula to solve the quadratic equation.

Quadratic Formula

$x=(-b \pm √(b^2-4ac))/(2a)\quad\textsf{when}\:ax^2+bx+c=0$

$\implies x=(-(-19)\pm √((-19)^2-4(3)(22)))/(2(3))$

$\implies x=(19\pm √(97))/(6)$

To find which value of x maximizes the volume, find the second derivative:

$\implies (d^2V)/(dx^2)=24x-76$

Then input the values of x into the second derivative:

$x=(19+√(97))/(6) \implies (d^2V)/(dx^2)=4√(97) > 0\implies \textsf{minimum}$

$x=(19-√(97))/(6) \implies (d^2V)/(dx^2)=-4√(97) < 0\implies \textsf{maximum}$

Therefore, the value of x that will maximize the volume is:

$x=(19-√(97))/(6)=1.53\:\sf inches \:(nearest\:hundredth)$

Alternatively, you can input the given options of x into the formula and compare results, but this is the correct way to find/prove it.

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