88496
The cost and revenue functions of a product are given by \(c(x)=20 x+4000\) and \(R(x)=60 x+\) 2000 respectively where \(x\) is the number of items produced and sold. The value of \(x\) to earn profit is
1 \(>50\)
2 \(>60\)
3 \(>80\)
4 \(>40\)
Explanation:
(A) : Given, \(\operatorname{cost} \mathrm{c}(\mathrm{x})=20 \mathrm{x}+4000\) and, Revenue \(\mathrm{R}(\mathrm{x})=60 \mathrm{x}+2000\) We know, Profit \(=\) Revenue - cost \(=(60 \mathrm{x}+2000)-(20 \mathrm{x}+4000)=40 \mathrm{x}-2000\) Hence, the value of \(x\) to earn some profit is \(40 \mathrm{x}-2000>0\) \(\mathrm{x}>50\)
Karnataka CET-2021
Linear Inequalities and Linear Programming
88497
The maximum value of \(P=6 x+8 y\), if \(\mathbf{2 x}+\mathbf{y} \leq \mathbf{3 0}, \mathrm{x}+\mathbf{2 y} \leq \mathbf{2 4 ;} \mathbf{x} \geq \mathbf{0}, y \geq \mathbf{0}\), will be
88498
If the total cost \(C(x)\) in rupees associated with the production of \(x\) units of an item is given by \(C(x)=3 x^{3}-2 x^{2}+x+100\). Then, the marginal change in cost, when \(x=5\), is
88496
The cost and revenue functions of a product are given by \(c(x)=20 x+4000\) and \(R(x)=60 x+\) 2000 respectively where \(x\) is the number of items produced and sold. The value of \(x\) to earn profit is
1 \(>50\)
2 \(>60\)
3 \(>80\)
4 \(>40\)
Explanation:
(A) : Given, \(\operatorname{cost} \mathrm{c}(\mathrm{x})=20 \mathrm{x}+4000\) and, Revenue \(\mathrm{R}(\mathrm{x})=60 \mathrm{x}+2000\) We know, Profit \(=\) Revenue - cost \(=(60 \mathrm{x}+2000)-(20 \mathrm{x}+4000)=40 \mathrm{x}-2000\) Hence, the value of \(x\) to earn some profit is \(40 \mathrm{x}-2000>0\) \(\mathrm{x}>50\)
Karnataka CET-2021
Linear Inequalities and Linear Programming
88497
The maximum value of \(P=6 x+8 y\), if \(\mathbf{2 x}+\mathbf{y} \leq \mathbf{3 0}, \mathrm{x}+\mathbf{2 y} \leq \mathbf{2 4 ;} \mathbf{x} \geq \mathbf{0}, y \geq \mathbf{0}\), will be
88498
If the total cost \(C(x)\) in rupees associated with the production of \(x\) units of an item is given by \(C(x)=3 x^{3}-2 x^{2}+x+100\). Then, the marginal change in cost, when \(x=5\), is
88496
The cost and revenue functions of a product are given by \(c(x)=20 x+4000\) and \(R(x)=60 x+\) 2000 respectively where \(x\) is the number of items produced and sold. The value of \(x\) to earn profit is
1 \(>50\)
2 \(>60\)
3 \(>80\)
4 \(>40\)
Explanation:
(A) : Given, \(\operatorname{cost} \mathrm{c}(\mathrm{x})=20 \mathrm{x}+4000\) and, Revenue \(\mathrm{R}(\mathrm{x})=60 \mathrm{x}+2000\) We know, Profit \(=\) Revenue - cost \(=(60 \mathrm{x}+2000)-(20 \mathrm{x}+4000)=40 \mathrm{x}-2000\) Hence, the value of \(x\) to earn some profit is \(40 \mathrm{x}-2000>0\) \(\mathrm{x}>50\)
Karnataka CET-2021
Linear Inequalities and Linear Programming
88497
The maximum value of \(P=6 x+8 y\), if \(\mathbf{2 x}+\mathbf{y} \leq \mathbf{3 0}, \mathrm{x}+\mathbf{2 y} \leq \mathbf{2 4 ;} \mathbf{x} \geq \mathbf{0}, y \geq \mathbf{0}\), will be
88498
If the total cost \(C(x)\) in rupees associated with the production of \(x\) units of an item is given by \(C(x)=3 x^{3}-2 x^{2}+x+100\). Then, the marginal change in cost, when \(x=5\), is
88496
The cost and revenue functions of a product are given by \(c(x)=20 x+4000\) and \(R(x)=60 x+\) 2000 respectively where \(x\) is the number of items produced and sold. The value of \(x\) to earn profit is
1 \(>50\)
2 \(>60\)
3 \(>80\)
4 \(>40\)
Explanation:
(A) : Given, \(\operatorname{cost} \mathrm{c}(\mathrm{x})=20 \mathrm{x}+4000\) and, Revenue \(\mathrm{R}(\mathrm{x})=60 \mathrm{x}+2000\) We know, Profit \(=\) Revenue - cost \(=(60 \mathrm{x}+2000)-(20 \mathrm{x}+4000)=40 \mathrm{x}-2000\) Hence, the value of \(x\) to earn some profit is \(40 \mathrm{x}-2000>0\) \(\mathrm{x}>50\)
Karnataka CET-2021
Linear Inequalities and Linear Programming
88497
The maximum value of \(P=6 x+8 y\), if \(\mathbf{2 x}+\mathbf{y} \leq \mathbf{3 0}, \mathrm{x}+\mathbf{2 y} \leq \mathbf{2 4 ;} \mathbf{x} \geq \mathbf{0}, y \geq \mathbf{0}\), will be
88498
If the total cost \(C(x)\) in rupees associated with the production of \(x\) units of an item is given by \(C(x)=3 x^{3}-2 x^{2}+x+100\). Then, the marginal change in cost, when \(x=5\), is
