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# h3mapr1 Maximizing Volume
Precalculus; Derivative (อนุพันธ์ของฟังก์ชัน)
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Owner wants to send some expensive fruit to overseas.
He gets the sample piece of carton paper size 10 cm. x 10 cm. from overseas customers.1. The owner orders the employee to fold this paper to be the box with an open top. 2. By cutting the paper sample according to the "X" tracing as the photo below. 3. Pack an expensive fruit inside the carton box with the open top of the transparent plastic in order to be able seeing through by customers. The worker must find the "X" value in this folding. When the size of paper is 10 cm.x 10 cm., 1. How many cm. of "X" value would be cut for the maximum volume of this carton box forming?. 2. What is the max. volume cm. cube of this carton box with open top?
 Solution | |||||
| Give "X" cm. is the length of the paper side which will be cut. And V(X) is the volume of the the carton box with the open top To value X>0 and 10-2X>0 ,therefore, 0 | |||||
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implies:
| V(X) |
V'(X)
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Conclusion
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X<5
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0
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Positive real number
| Increasing function |
| X=5/3 |
0
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Zero
| Relative maximum value |
| X>5/3 |
0
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Negative real number
| Decreasing function |
 |
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Test by second derivative
The second derivative is negative. The function therefore has a relative maximum
value at x = 5/3
Ans. Therefore at X = 5/3 cm., the carton box open top will have a max. volume.
The max. volume cm. cube of this carton box with open top is 74.074 cm. cubic.
To the left of a maximum -- at a point C -- the slope of the tangent is positive: f '(x) > 0.
While at points immediately to the right -- at a point D -- the slope is negative: f '(x) < 0.
In other words, at a maximum, f '(x) changes sign from + to −
