a box with a square base and open top must have a volume of 78732 cm³. we wish to find the dimensions of the…

a box with a square base and open top must have a volume of 78732 cm³. we wish to find the dimensions of the box that minimize the amount of material used. first, find a formula for the surface area of the box in terms of only x, the length of one side of the square base. hint: use the volume formula to express the height of the box in terms of x. simplify your formula as much as possible. a(x)= next, find the derivative, a(x). a(x)= now, calculate when the derivative equals zero, that is, when a(x)=0. hint: multiply both sides by x². a(x)=0 when x= we next have to make sure that this value of x gives a minimum value for the surface area. lets use the second - derivative test. find a(x). a(x)= evaluate a(x) at the x - value you gave above. note: since your last answer is positive, this means that the graph of a(x) is concave up around that value, so the zero of a(x) must indicate a local minimum for a(x). (your boss is happy now.)

a box with a square base and open top must have a volume of 78732 cm³. we wish to find the dimensions of the box that minimize the amount of material used. first, find a formula for the surface area of the box in terms of only x, the length of one side of the square base. hint: use the volume formula to express the height of the box in terms of x. simplify your formula as much as possible. a(x)= next, find the derivative, a(x). a(x)= now, calculate when the derivative equals zero, that is, when a(x)=0. hint: multiply both sides by x². a(x)=0 when x= we next have to make sure that this value of x gives a minimum value for the surface area. lets use the second - derivative test. find a(x). a(x)= evaluate a(x) at the x - value you gave above. note: since your last answer is positive, this means that the graph of a(x) is concave up around that value, so the zero of a(x) must indicate a local minimum for a(x). (your boss is happy now.)

Answer

Explanation:

Step1: Find height in terms of x

Let the side - length of the square base be $x$ and the height of the box be $h$. The volume $V=x^{2}h$. Given $V = 78732$, then $h=\frac{78732}{x^{2}}$.

Step2: Find surface - area formula

The surface area $A(x)$ of an open - top box with a square base is $A(x)=x^{2}+4xh$. Substitute $h=\frac{78732}{x^{2}}$ into the surface - area formula: $A(x)=x^{2}+4x\cdot\frac{78732}{x^{2}}=x^{2}+\frac{314928}{x}$.

Step3: Find the first derivative

Using the power rule, if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. For $A(x)=x^{2}+\frac{314928}{x}=x^{2}+314928x^{-1}$, $A^\prime(x)=2x-314928x^{-2}=2x-\frac{314928}{x^{2}}$.

Step4: Find when the first derivative is zero

Set $A^\prime(x) = 0$, so $2x-\frac{314928}{x^{2}}=0$. Multiply both sides by $x^{2}$: $2x^{3}-314928 = 0$. Then $x^{3}=\frac{314928}{2}=157464$, and $x = 54$.

Step5: Find the second derivative

Differentiate $A^\prime(x)=2x - 314928x^{-2}$ with respect to $x$. $A^{\prime\prime}(x)=2 + 629856x^{-3}=2+\frac{629856}{x^{3}}$.

Step6: Evaluate the second derivative at $x = 54$

Substitute $x = 54$ into $A^{\prime\prime}(x)$: $A^{\prime\prime}(54)=2+\frac{629856}{54^{3}}=2+\frac{629856}{157464}=2 + 4=6$.

Answer:

$A(x)=x^{2}+\frac{314928}{x}$ $A^\prime(x)=2x-\frac{314928}{x^{2}}$ $x = 54$ $A^{\prime\prime}(x)=2+\frac{629856}{x^{3}}$ $A^{\prime\prime}(54)=6$