A box with a square base and open top must have a volume of 131072 cm^3. 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 x^2.]A′(x)=0 when x=We next have to make sure that this value of gives a minimum value for the surface area. Let's 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 131072 cm^3. 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 x^2.]A′(x)=0 when x=We next have to make sure that this value of gives a minimum value for the surface area. Let's 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 131072 cm^3. 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 x^2.]A′(x)=0 when x=We next have to make sure that this value of gives a minimum value for the surface area. Let's 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 131072 cm^3. 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 x^2.] A′(x)=0 when x=
We next have to make sure that this value of gives a minimum value for the surface area. Let's 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.)
Formula Formula A function f ( x ) is also said to have attained a local minimum at x = a , if there exists a neighborhood ( a − δ , a + δ ) of a such that, f ( x ) > f ( a ) , ∀ x ∈ ( a − δ , a + δ ) , x ≠ a f ( x ) − f ( a ) > 0 , ∀ x ∈ ( a − δ , a + δ ) , x ≠ a In such a case f ( a ) is called the local minimum value of f ( x ) at x = a .
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