There is no maximum since a quartic with a positive leading coefficient increases without limit.
The minimum is approximately -4.4940
I'm going to assume you mean F(x) = -2x2 + 10x + 6 as -2x plus 10x = 8x...First differentiate -2x2 + 10x + 6:dy/dx = -4x + 10Set dy/dx = 0:-4x + 10 = 0Find x:10 = 4x2.5 = xFind y at x =2.5:y = -2(2.5)2 + 10(2.5) + 6 = -12.5 + 25 +6 = 18.5So max or min is at (2.5, 18.5)To find out whether it is max or min differentiate twice:d2y/dx2 = -4If its negative, point will be a maximumIf its positive, point will be a minimum.In this case, a maximum occurs at (2.5, 18.5).
take the max plus the min and divide the anser by to
#include <stdio.h> #include <stdlib.h> #include <time.h> int GetRand( int min, int max) { // Swap min and max if wrong way around. if(max<min)min^=max^=min^=max; return(rand()%(max-min+1)+min); } int main() { // Seed the random number generator with current time to ensure // random numbers are different on each run. // Note: use srand(1) to generate the same sequence on every run. srand((unsigned)time(NULL)); printf("Number from 0 to 10 : %d\n", GetRand(0,10)); printf("Number from 1 to 100: %d\n", GetRand(1,100)); return(0); }
=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)
Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. sonika aggarwal GNIIT
Use the median-of-three algorithm: int min (int a, int b) { return a<b?a:b; } int max (int a, int b) { return a<b?b:a; } int median_of_three (int a, int b, int c) { return max (min (a, b), min (max (a, b), c)); } Note that the algorithm does not cater for equal values which creates a problem when any two values are equal, because there are only two values to play with, neither of which can be regarded as being the middle value. If the equal value is the lower of the two values, the largest value is returned if and only if it is the last of the three values, otherwise the lowest value is returned. But when the equal value is the larger of the two values, the largest value is always returned. Lowest value is equal: Input: 0, 0, 1 = max (min (0, 0), min (max (0, 0), 1)) = max (0, min (0, 1)) = max (0, 1) = 1 Input: 0, 1, 0 = max (min (0, 1), min (max (0, 1), 0)) = max (0, min (1, 0)) = max (0, 0) = 0 Input: 1, 0, 0 = max (min (1, 0), min (max (1, 0), 0)) = max (0, min (1, 0)) = max (0, 0) = 0 Highest value is equal: Input: 0, 1, 1 = max (min (0, 1), min (max (0, 1), 1)) = max (0, min (1, 1)) = max (0, 1) = 1 Input: 1, 0, 1 = max (min (1, 0), min (max (1, 0), 1)) = max (0, min (1, 1)) = max (0, 1) = 1 Input: 1, 1, 0 = max (min (1, 1), min (max (1, 1), 0)) = max (1, min (1, 0)) = max (1, 0) = 1 The only way to resolve this problem and produce a consistent result is to sum all three inputs then subtract the minimum and maximum values: int median_of_three (int a, int b, int c) { return a + b + c - min (min (a, b), c) - max (max (a, b), c)); } Lowest value is equal: Input: 0, 0, 1 = 0 + 0 + 1 - min (min (0, 0), 1) - max (max (0, 0), 1) = 1 - 0 - 1 = 0 Input: 0, 1, 0 = 0 + 1 + 0 - min (min (0, 1), 0) - max (max (0, 1), 0) = 1 - 0 - 1 = 0 Input: 1, 0, 0 = 1 + 0 + 0 - min (min (1, 0), 0) - max (max (1, 0), 0) = 1 - 0 - 1 = 0 Highest value is equal: Input: 0, 1, 1 = 0 + 1 + 1 - min (min (0, 1), 1) - max (max (0, 1), 1) = 2 - 0 - 1 = 1 Input: 1, 0, 1 = 1 + 0 + 1 - min (min (1, 0), 1) - max (max (1, 0), 1) = 2 - 0 - 1 = 1 Input: 1, 1, 0 = 1 + 1 + 0 - min (min (1, 1), 0) - max (max (1, 1), 0) = 2 - 0 - 1 = 1 This makes sense because when we sort 0, 0, 1 in ascending order, 0 is in the middle, while 0, 1, 1 puts 1 in the middle.
int sum (int min, int max) {return (max-min+1)*(max+min)/2;}
The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)
int a, b, c, d, max, min; scanf("%d%d%d%d",&a, &b, &c, &d); (a>b)?(max=a,min=b):(max=b,min=a); (c>d)?(a=c,b=d):(a=d,b=c); max=(a>max)?a:max; min=(b<min)?b:min; printf("%d %d\n", max, min);
min: 0.5 KVA MAX: 1.5 KVA
Usually, max 212 F and min 32 F It depends on for what the thermometer will be used. My greenhouse min/max thermometer has a min of -40°F and a max of 120°F A (non-digital) clinical thermometer has a min of 94°F and a max of 108°F A cook's (sugar/jam) thermometer has a min of 100°F and a max of 400°F
If all the readings are stored in an array, then the question becomes how to determine the maximum and minimum values in the array. This can be achieved with the following function: double min(double a[], const size_t size) { double min=a[0]; for( size_t i=1; i<size; ++i ) if(a[i]<min) min=a[i]; return(min); } double max(double a[], const size_t size) { double max=a[0]; for( size_t i=1; i<size; ++i ) if(a[i]<max) max=a[i]; return(max); } Note that when taking temperature readings, you will probably take several samples per day, possibly as often as once every second to obtain a high degree of accuracy. If so, your array will have 31.5 million elements each year.